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# How do you solve this system of equations: (x - 3) ^ { 2} + ( y + 2) ^

###### Question:

How do you solve this system of equations: (x - 3) ^ { 2} + ( y + 2) ^ { 2} = 16 and x - 2y = 10?

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# Thread: Present/Future Value of Annuity Question

1. ## Present/Future Value of Annuity Question

Jane Doe is 25 years old and starting next year wants to
save and invest in order to ensure an annual income of $100kin each of the 20 years after she retires, till she is 85. She willretire at 65. She expects to earn 10% interest till retirement and 5%interest in the years after retirement. How much must she save each year from now till retirementstarting next year? This part is easy; PVA with pmt=100k, t=20, r=5% --> ~$1,246,200

The second part: As it turned out, Jane spent the next 5 years on the beach inBarcelona. She is 30 now and has $4,000 in the bank. Howmuch does she need to save each year, starting next year?All other data unchanged. What should be my thought process behind this? I know the future value when she's 65 should be$1,246,200, interest should be 10%, and there's 35 more years.

2. ## Re: Present/Future Value of Annuity Question

Hey youngb11.

What formulas do you have for the annuity?

What you can do is either re-arrange to solve for n or use a numerical scheme if you can't re-arrange n by itself on only one side of the equation.

3. ## Re: Present/Future Value of Annuity Question

Originally Posted by youngb11
Jane Doe is 25 years old and starting next year wants to
save and invest in order to ensure an annual income of $100kin each of the 20 years after she retires, till she is 85. She willretire at 65. She expects to earn 10% interest till retirement and 5%interest in the years after retirement. How much must she save each year from now till retirementstarting next year? This part is easy; PVA with pmt=100k, t=20, r=5% --> ~$1,246,200

The second part: As it turned out, Jane spent the next 5 years on the beach inBarcelona. She is 30 now and has $4,000 in the bank. Howmuch does she need to save each year, starting next year?All other data unchanged. What should be my thought process behind this? I know the future value when she's 65 should be$1,246,200, interest should be 10%, and there's 35 more years.
I would give it a try, though I have had limited exposure to business math...I will do both the problems as I see that you have partly done the first problem and maybe it'll be more useful to everybody else...

1. Jane wants $1000 per year after retirement at 5% interest, so she would want to know the present value of annuity(pva) at the point of retirement,$\displaystyle pva=p*\frac{1-\frac{1}{(1+r)^n}}{r}$#p=>pension & r=>interest rate & n=> no. of years putting the values,$\displaystyle pva=100000*\frac{1-\frac{1}{(1+0.5)^{20}}}{0.5}=1246221$Now that jane has to have$ 1246221 at retirement, it becomes a future value problem now:
$\displaystyle fv=c*\frac{(1+r)^n - 1}{r}$ # c=cash payments to bank
since jane wants to have 1246221 as future value at retirement,
$\displaystyle 1246221=c*\frac{(1+0.1)^{40} - 1}{0.1}$
therefore, $\displaystyle c=2815.72$
so jane would need to save at least $\displaystyle \$ 2815.72$and invest it, to fulfill her plans... 2. I guess the$4000 accumulated by jane would get compounded annually and contribute in making the $1246221, which she requires at the point of retirement.$\displaystyle A=p(1+r)^n=4000(1+0.1)^{35}=112409.74$now we will again calculate c (cash payments to bank) like we did before but take into account the$ 1246221
$\displaystyle fv=1246221=112409.74 + c\frac{(1+0.1)^{35} -1}{0.1}$
$\displaystyle =1133811.26=c*\frac{(1+0.1)^{35} -1}{0.1}$
solving for c we get $\displaystyle c=\$ 4183.49$note that in the second case jane's savings or cash payment to bank has increased from$2815.72 to $4183.49. This is because, Jane would have made more by saving what she had for 5 years than she had earned at barcelona.If she had saved$ 2815.72 per year for 5 years, her account would have gone to $\displaystyle 2815.72*\frac{(1+0.1)^5 -1}{0.1}$ or
$\displaystyle \$ 17190.25 \$.

So jane made a terrible decision and I have tried to make approximations upto 2 decimal places
please correct me if i am wrong.

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Important Questions for Class 12 Physics Chapter 6 Electromagnetic Induction Class 12 Important Questions

# Important Questions for Class 12 Physics Chapter 6 Electromagnetic Induction Class 12 Important Questions

## Important Questions for Class 12 Physics Chapter 6 Electromagnetic Induction Class 12 Important Questions

#### Electromagnetic Induction Class 12 Important Questions Very Short Answer Type

Question 1.
A plot of magnetic flux (ϕ ) versus current (I) is shown in the figure for two inductors A and B. Which of the two has larger value of self inductance? (Delhi 2010)

Since ϕ = LI
∴ L = $$\frac{\phi}{I}$$ = slope
Slope of A is greater than slope of B
∴ Inductor A has larger value of self inductance than inductor B.

Question 2.
Define self-inductance of a coil. Write its S.I. unit. (All India 2010)
Self induction is the property of a coil by virtue of which it opposes the growth or decay of the current flowing through it.
S.I. unit of self-inductance is henry (H).

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Question 3.
Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor. (All India 2013)

When both magnets move towards loop, the A side plate of cL capacitor will be positive while the lower plate B is negative, making the induced current in a clockwise direction.

Question 4.
Predict the polarity of the capacitor when the two magnets are quickly moved in the directions marked by arrows.

Curerent in the coil will be anti-clockwise, when seen from the left, therefore plate A will become + ve (positive) and plate B will be negative.

Question 5.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.

In metal ring 1, the induced current flows in the clockwise direction.
In metal ring 2, the induced current flows in the anticlockwise direction.

Question 6.
Predict the direction of induced current in a metal ring when the ring is moved towards a conductor is carrying current I in the direction shown in the figure.

Clockwise direction.

Question 7.
Predict the directions of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing?

In metal ring 1, the induced current flows in Anticlockwise direction.
In metal ring 2, the induced current flows in the Clockwise direction.

Question 8.
A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the directions of induced current in each coil. (All India 2012)

By Lenz’s law, the ends of both the coils closer to the magnet behave as south pole. Hence the current induced in both the coils will flow clockwise when seen from the magnet side.

Question 9.
State Lenz’s law. (Comptt. All India 2012)
Lenz’s law states that “the polarity of induced emf is such that it tends to produce a current, which opposes the change in magnetic flux that induced it”.

Question 10.
Predict the direction of the induced current in the rectangular loop abed as it is moved into the region of a uniform magnetic field $$\vec{B}$$ directed normal to the plane of the loop. (Comptt. All India 2012)

The direction of the induced current in the given rectangular loop is anti-clockwise, i.e., cbadc.

Question 11.
State Faraday’s law of electromagnetic induction. (Comptt. All India 2012)
Faraday’s law states that “The magnitude of emf induced in a circuit is directly proportional to the rate of change of magnetic flux linked with the circuit”. Mathematically, we can write,
$$e=-\frac{d \phi}{d t}$$ …where [dϕ is the small change in magnetic flux in small time dt

Question 12.
How does the mutual inductance of a pair of coils change when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased (All India 2013)
(i) Mutual inductance decreases’, because flux linked with the secondary coil decreases.
(ii) $$\mathrm{M}=\mu_{0} n_{1} n_{2} \mathrm{Al}$$, so when n1 and n2 increase, mutual inductance (M) increases.

Question 13.
A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason. (All India 2013)
Because of Eddy Current

If the upper force of the core of the electromagnet acquires north polarity, then according to Lenz’s Law, the lower face of the disc will also acquire north polarity. Due to the force of repulsion between the lower face (N-pole) of the core of the electromagnet, the disc jumps upto a certain height.

Question 14.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What do the cause of this damping? (All India 2013)
The cause of this damping is eddy current.

Question 15.
The motion of copper plates is damped when it is allowed to oscillate between the two poles of a magnet. If slots are cut in the plate, how will the damping be affected? (All India 2013)
Eddy current will decrease due to which damping reduces.

Question 16.
How does the mutual inductance of a pair of coils change when
(i) distance between the coils is decreased and
(ii) number of turns in the coils is decreased? (All India 2013)
(i) increases.
(ii) decreases, because
$$\mathbf{M}=m_{0} n_{1} n_{2} \mathrm{Al}$$
where [N1 and N2 are number of turns

Question 17.
Predict the polarity of the capacitor in the situation described in the figure. (Comptt. Delhi 2013)

When both magnets move towards loop, the A side plate of cL capacitor will be positive while the lower plate B is negative, making the induced current in a clockwise direction.

Question 18.
Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why? (Delhi 2014)
Glass bob would reach earlier because there would be a force acting upward due to eddy currents on metallic bob being conducting, due to earth’s magnetic field. This will slow down the metallic bob.

Question 19.
The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. (All India 2014)

The direction of current in loop wire will be clockwise.

Question 20.
A conducting loop is held above a current carrying figure. Depict the direction of the current induced in the loop when the current in the wire PQ is constantly increasing. (All India 2014)

The current induced is in clockwise direction.

Question 21.
A conducting loop is held below a P current carrying wire PQ as shown. Predict the direction of the induced current in the loop when the current in the wire is constantly increasing. (All India 2014)

The current induced is in anti clockwise direction.

Question 22.
A metallic piece gets hot when surrounded by a coil carrying high frequency alternating current. Why? (Comptt. Delhi 2014)
Due to the heating effect of eddy currents set up in the metallic piece.

Question 23.
Predict the polarity of the plate A of the capacitor, when a magnet is moved ff’ towards it, as shown in the figure. (Comptt. All India)

The plate ’A’ has the positive polarity.

Question 24.
Define the term ‘self-inductance’ of a coil. Write its S.I. Unit. (All India 2015)
The self inductance of a coil numerically equals to the induced emf produced in the coil, when the rate of change of current in the coil is unity.

(where L is the self inductance of the coil)
S.I. Unit : Henry.

Question 25.
Name any two applications where eddy currents are used to advantage. (Comptt. Delhi 2015 )
Applications of Eddy currents :

1. Electromagnetic Damping
2. Magnetic Breaking
3. Induction Furnace
4. Electric Power metres (any two)

Question 26.
A long straight current carrying wire passes normally through the centre of a circular loop. If the current through the wire increases, will there be an induced emf in the. loop? Justify. (Delhi 2015)
No,As the magnetic field due to current carrying wire will be in the plane of the circular loop, so magnetic flux will remain zero.

Question 27.
Predict the polarity of the capacitor in the situation described in the given diagram. (All India 2017)

The polarity of plate ‘A’ of the given capacitor is positive, while that of plate ‘B’ is negative.

Question 28.
A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the direction of the induced current in each coil (All India 2017)

Induced current flows from P to Q through ammeter;
while it flows from D to C through ammeter.

Question 29.
What is the direction of induced currents in metal rings 1 and 2 when current I in the wire is increasing steadily? (All India 2017)

The direction of induced current is clockwise in metal ring ‘1’ and anti-clockwise in metal ring ‘2’.

Question 30.
In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. (Comptt. All India 2017)

The polarity of plate is positive; 1 while that of plate B is negative.

#### Electromagnetic Induction Class 12 Important Questions Short Answer Type-II

Question 31.
Derive an expression for the self-inductance of a long air-cored solenoid of length l and number of turns N. (Delhi 2008)
Consider a long solenoid of length l and radius r with r << 1 and having n turns per unit length. If a current I flows through the coil, then the magnetic field inside the coil is almost constant and is given by
B = µ0nI
Magnetic flux linked with each turn = BA = µ0nIA

Question 32.
(i) When primary coil P is moved towards secondary coil S (as shown in the figure) the galvanometer shows momentary deflection. What can be done to have larger deflection in the galvanometer with the same battery?

(ii) State the related law. (Delhi 2010)
(i) To have larger deflection in the galvanometer with the same battery, coil P has to be moved faster towards S so that rate of change of magnetic flux is more.

(ii) The related law governing this phenomenon is Faraday’s second law of electromagnetic induction which states that induced emf is set up in a circuit when magnetic flux linked with it changes. The magnitude of induced emf is proportional to the rate of change of magnetic flux.

Question 33.
A coil Q is connected to low voltage bulb B and placed near another coil P as shown in the figure. Give reasons to explain the following observations :

(a) The bulb ‘B’ lights.
(b) Bulb gets dimmer if the coil Q is moved towards left. (Delhi 2010)

1. The bulb B lights on account of emf induced in the coil Q due to mutual induction between P and Q.
2. When coil Q is moved towards left, magnetic flux linked with Q decreases and may even reduce to zero at some distance. The emf induced may decrease and the bulb B gets dimmer.

Question 34.
Two identical loops, one of copper and the other of aluminium, are rotated with the same angular speed in the same magnetic field. Compare
(i) the induced emf and
(ii) the current produced in the two coils. Justify your answer.(All India 2010)
(i) Induced emf in a coil is ε = NBAω sin ωt
As the angular speed is same, induced emf will also be same in both the loops.
(ii) Current induced in a loop is $$I=\frac{\varepsilon}{R}=\frac{\varepsilon A}{\rho l}$$
As the resistivity of copper is lesser, more amount of current is induced in it.

Question 35.
A rectangular loop and a circular loop are moving out of a uniform magnetic field to a field-free region with a constant velocity ‘v’ as shown in the figure. Explain in which loop do you expect the induced emf to be constant during the passage out of the field region. The magnetic field is normal to the loops. (All India 2010)

In case of rectangular loop, induced emf will be constant. It is because rate of change of the area of rectangular loop is uniform whereas that of a circular loop is not constant.

Question 36.
A current is induced in coil C1 due to the motion of current carrying coil C2.

(a) Write any two ways by which a large deflection can be obtained in the galvanometer G.
(b) Suggest an alternative device to demonstrate the induced current in place of a galvanometer .
(a) To obtain a large deflection, one or more of the following steps can be taken :

1. Use a rod made of soft iron inside the coil C2.
2. Connect the coil to a powerful battery.
3. Move the arrangement rapidly towards the test coil C1

(b) Replace the galvanometer by a small torch bulb.

Question 37.
A magnet is quickly moved in the direction indicated by an arrow between two coils C1 and C2 as shown in the figure. What will be the direction of induced current in each coil as seen from the magnet? Justify your answer. (Delhi 2011)

When magnet moves in given direction, induced current will be clockwise in both the coils as magnet is going away from C1 and moving towards C2 making C1 as S pole and C2 also as S pole according to Lenz rule. So C1 will try to attract and C2 will try to repel the motion of magnet.

Question 38.
What are eddy currents? Write any two applications of eddy currents. (All India 2011)
(a) Eddy current : Due to change in magnetic flux, if there is induced current in the volume (bulk) of the material, it is called as eddy currents. It is a necessary evil in an arrangement as it can be used in applications like electric brakes, induction furnaces and dead-beat galvanometers and brings loss of energy with heat production etc.
(b) (i) As the arm RS of length l is moved with a uniform speed, there is a change in area. It is given by dA = Idx = Ivdt

Question 39.
Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self inductance L is given by – $$\frac{1}{2}$$I2. (Delhi 2011)
The self-inductance of a coil may be defined as the induced emf set up in the coil due to a unit rate of change of current through it.
Let I be the current through the inductor L at any instant t
The current rises at the rate dlldt, so the induced

Work done against the induced emf in small time dt is

Total work done in building up the current from 0 to I in

This work done is stored as the magnetic field energy U in the inductor

Question 40.
Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual inductance of inner solenoid of length l having the radius r1 and the number of turns n1 per unit length due to the second outer solenoid of same length and n2 number of turns per unit length. (Delhi 2011)
Mutual induction is the phenomenon of inducing emf in a coil due to the rate of change of current in a nearby coil.

When a current I2 is set up through S2 (outer solenoid), it in turn sets up a magnetic flux through S1. Let us denote it by ϕ 1, The corresponding flux linkage with solenoid S1 is,

(Here N2 is total no. of turns with S2)
M21 is referred to as coefficient of mutual inductance of solenoid S2 with respect to solenoid S1.
Thus, flux linkage with solenoid S2 is

Using equations (2) and (3), we get
M12 = M21 = M
If a medium of relative permeability (it had been present the mutual inductance would be

Question 41.
A metallic rod of ‘L’ length is rotated with angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. (Delhi 2012)
The magnitude of the emf, generated across a length dr of the rod, as it moves at right angles to the magnetic field, is given by
dε = Bvdr

Alternatively, The potential difference across the resistor is equal to the induced emf and equal B x .(rate of change of area of loop), If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ (as shown in the figure) is given by

where [R is the radius of the circle]

Question 42.
Derive the expression for the self inductance of a long solenoid of cross sectional area A and. length l, having n turns per unit length.(Delhi 2012)
Self-induction of a long solenoid : Consider a long solenoid of length l and radius r with r << 1 and having n turns per unit length. If a current I flows through the coil, then the magnetic field inside the coil is almost constant and is given by

Magnetic flux linked with each turn,

Question 43.
State Lenz’s Law.
A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. (Delhi 2012)
Lenz’s law states that “the polarity of induced emf is such that it tends to produce a current, which oppose the change in magnetic flux that induced it”.

Yes there will be an emf induced as the horizontal component of field of earth, velocity of the motion of the rod and the length of the rod are all perpendicular to each other.

The magnetic flux due to vertical component of Earth’s magnetic field keeps on changing as the metallic rod falls down.

Question 44.
A rectangular loop PQMN with movable arm PQ of length 10 cm and resistance 2 Ω is placed in a uniform magnetic field of 0.1 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQ are negligible. Calculate the
(i) emf induced in the arm PQ and
(ii) current induced in the loop when arm PQ is moved with velocity 20 m/s. (Comptt. Delhi 2014)

(i) emf induced

(ii) Current in the loop,

Question 45.
A rectangular loop PQMN with movable arm PQ of length 20 cm and resistance 5 Ω is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the loop as is shown in the figure.

The resistances of the arms MN, NP and MQ are negligible. Calculate the
(i) emf induced in the arm PQ and
(ii) current induced in the loop when arm PQ is moved with velocity 15m/s (Comptt. Delhi 2014)

#### Electromagnetic Induction Class 12 Important Questions Short Answer Type – II SA – II

Question 46.
A metallic rod of length l is rotated at a constant angular speed to, normal to a uniform magnetic field B. Derive an expression for the current induced in the rod, if the resistance of the rod is R. (Delhi 2008)
emf produced/induced across a small-section (dr) on the rod

Question 47.
An inductor 200 mH, capacitor 500 µF, resistor 10 Ω are connected in series with a 100 V variable frequency a.c. source. Calculate the
(i) frequency at which the power factor of the circuit is unity
(ii) current amplitude at this frequency
(iii) Q-factor (Delhi 2008)
(i) Power factor will be unity at resonance,

Question 48.
A coil of number of turns N, area A, is rotated at a constant angular speed ω, in a uniform magnetic field B, and connected to a resistor R. Deduce expressions for :
(i) Maximum emf induced in the coil.
(ii) Power dissipation in the coil. (Delhi 2008)
We know that induced emf

Question 49.
(a) Define self inductance. Write its S.I. units.
(b) Derive an expression for self inductance of a long solenoid of length l, cross-sectional area A having N number of turns. (Delhi 2009)
(a) Self-inductance of a coil
Since flux ϕ = LI

where [L is coefficient of self-induction or self inductance] Self inductance is numerically equal to the magnetic flux linked with the coil when unit current passes through it.
Its S.I. unit is henry.
(b)
Consider a long solenoid of length l and radius r with r << 1 and having n turns per unit length. If a current I flows through the coil, then the magnetic field inside the coil is almost constant and is given by
B = µ0nI
Magnetic flux linked with each turn = BA = µ0nIA

Question 50.
(i) State Faraday’s law of electromagnetic induction.
(ii) A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10-4 T and the dip angle is 30°? (All India)
(i) First law : Whenever the magnetic flux linked with a closed circuit changes, an emf (and hence a current) is induced in it which lasts only so long as the change in flux is taking place. This phenomenon is called electromagnetic induction.
Second law : The magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the closed circuit.

Since v, l and B are to be perpendicular to get induce emf, the vertical component alone will contribute to the induction

– ve sign signifies that the emf will oppose the change in magnetic flux causing it.

Question 51.
State the law that gives the polarity of the induced emf. (All India 2009)
Law that gives the polarity of the induced emf is :
Lenz’s law : “The direction of induced e.m.f. is such that it opposes the cause which produces it. The polarity of induced e.m.f. can be determined by Lenz’s law.”

Question 52.
A metallic rod of length ‘l’ is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present every where. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtain the expression for it. (Delhi 2013)
The magnitude of the emf, generated across a length dr of the rod, as it moves at right angles to the magnetic field, is given by
dε = Bvdr

Alternatively, The potential difference across the resistor is equal to the induced emf and equal B x .(rate of change of area of loop), If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ (as shown in the figure) is given by

where [R is the radius of the circle]

Question 53.
Starting from the expression for the energy w = $$\frac{1}{2} \mathbf{L I}^{2}$$, stored in a solenoid of self-inductance L to build up the current I, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and length l of the solenoid having n number of turns per unit length. Hence show that the energy density is given by B2/2µ0. (Comptt. Delhi 2013)

From these two equations, we get

Question 54.
A metallic rod of length ‘l’ is rotated with a uniform angular speed ω, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R = l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present eveywhere. Deduce the expression for the emf induced in the rod.
If r is the resistance of the rod and the metallic ring has negligible resistance, obtain the expression for the power generated. (Comptt. All India 2013)
Are swept per rotation = πR2 = πl2 [∵ R = l]

Question 55.
Write its SI unit for self-inductance of a coil. Derive the expression for self-inductance of a long solenoid of cross-sectional area ‘A’, length ‘l’ having ‘n’ turns per unit length. (Comptt. All India 2013)
Coefficient of self induction.
Consider a coil L as shown in Figure. Suppose a current I flows through the coil at any instant then magnetic flux ϕ linked with the coil is directly proportional to the current passing through it at that instant.

⇒ ϕ = LI where [L is called coefficient of self induction] If I = 1, then ϕ = L
Thus, self inductance of a coil is numerically equal to the magnetic flux linked with the coil, when a unit current flows through it.

The SI unit of self inductance is henry (H).

Expression for self-inductance : Consider a long solenoid of length Z and cross-sectional area A having n turns per unit length.
The magnetic field due to a current flowing in the solenoid is B = µ0nl
Total flux linked with the solenoid is,

where [nl is the total number of turns.] Thus, the self inductance is, $$\frac{\phi}{I}$$

If we fill the inside of the solenoid with a material of relative permeability µr then
L = µrµ0n2Al

Question 56.
A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. The Earth’s magnetic field at the place is 0.4 G and the angle of dip is 60°. Calculate the emf induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased? (All India 2013)
If a rod length ‘l’ rotates with angular speed ω in the uniform magnetic field B,

The induced emf will not change with the increase in the number of spokes.

Question 57.
Define the term ‘mutual inductance’ between the two coils.
Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r1 and r2 (r2 >> r1). Total number of turns in the two solenoids are N1 and N2 respectively. (All India 2013)
(i) Mutual inductance. The mutual inductance of two coils is numerically equal to the induced emf produced in one coil, when the rate of change of current is unit in the other coil.

(ii)

(a) Lenz’s law (Direction) : “The direction of induced current in a circuit is such as to oppose
the change that causes it. So $$e m f=-\frac{d \phi}{d t}$$
When north pole of a magnet, for example, is brought near a coil, then the nearer end of the coil will acquire north polarity and mechanical work will be needed to cause the relative motion between the coil and the magnet. This work, in fact, converts itself in the form of induced emf. Hence, the phenomenon of electromagnetic induction is in accordance with the law of conservation of energy.

(b) When a current I2 is set up through S2, it in turn sets up a magnetic flux through S1. Let us denote it by ϕ1. The corresponding flux linkage with solenoid S1 is

M12 is called the mutual inductance of solenoid S1 with respect to solenoid S2.
The magnetic field due to the current I2 in S2

We now consider the reverse case :
A current I1, is passed through the solenoid S1 and the flux linkage with coil S2 is,

M21 is called the mutual inductance of the solenoid S2 with respect to solenoid S1

The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoid are very long. Thus, flux linkage with solenoid S2 is

Question 58.
Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it. (All India 2014)
Self-inductance: emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as

where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using the above equation the induced emf is given by

Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil. It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross-sectional area A and length Z, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is B = µ0nI (neglecting edge effects, as before). The total flux linked with the solenoid is

If we fill the inside of the solenoid with a material of relative permeability (e.g. soft iron, which has a high value of relative permiability), then L = µrµ0n2Al

The self-inductance of the coil depends on its geometry and on the permeability of the medium. The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. Physically, the self-inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf (ε) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is

If we ignore the resistive losses and consider only inductive effect, then using

Total amount of work done in establishing the current I is,

Thus, the magnetic energy required to build up the current I is,

Question 59.
(a) A rod of length l is moved horizontally with a uniform velocity ‘v’ in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.
(b) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain. (All India 2014)
(a) Induced emf in rotating metallic rod: Suppose a rectangular loop LMNO is placed in a uniform magnetic field B

Suppose at any instant, length ON = x
Area of the loop LMNO = lx
Flux through the loop, ϕ = Blx

(b) The induced emf produced in a rod (l) moved with velocity (v) kept in a magnetic field (perpendicular to the plane of length of the rod) is given by e = BIv …(i)

This induced emf is called motional emf, by moving a conductor instead of varying the magnetic field; that is by changing the magnetic flux enclosed by the circuit.

We can explain motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor.
Lorentz force acting on charge q is

The work done in moving, the charge through a distance is : W = qvBl
Since the emf is the work done per unit

This expression is the same as given in (i)

Question 60.
Derive the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid carrying a steady current I. How does this magnetic energy per unit volume compare with the electrostatic energy density stored in a parallel plate capacitor? (Comptt. Delhi 2015)
Magnetic Energy in a Solenoid,

The electrostatic energy stored per unit volume for a parallel plate capacitor,

These two expressions are similiar in nature.

Question 61.
(i) Define mutual inductance.
(ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (Delhi 2015)
(i) Mutual inductance : Magnetic flux, linked with the secondary coil due to the unit current flowing in the primary coil, $$\phi_{2}=\mathrm{MI}_{1}$$
Therefore, induced emf associated with the secondary coil, for a unit rate of change of current in the primary coil. $$e_{2}=-\mathrm{M} \frac{d l_{1}}{d t}$$
[i.e. the phenomenon of production of induced emf in one coil due to change in current in neighbouring coil]

(ii) Given : M = 1.5 H, I1 = 0, I2 = 20 A, ∆t = 0.5 sec, dϕ = ?
Change in flux (dϕ) = MdI = M(I2 – I1) = 1.5 (20 – 0) = 30 Weber

Question 62.
(a) Define self-inductance of a coil and hence write the definition of ‘Henry’.
(b) Write any two factors each on which the following depends :
(i) Self inductance of a coil.
(ii) mutual inductance of a pair of coils. (Comptt. All India 2015)
(a) The self-inductance (L) of a coil equal the magnetic flux linked with it, when a unit current flows through it.
One henry is the self inductance of a coil for which the magnetic flux, linked with it, due to a current of 1A, flowing in it, equals one weber.

(b) (i) Self inductance of a coil depends on

• Its geometry (area and length of a coil.
• Number of turns
• Medium within the coil

(ii) Mutual inductance of a given pair of coils depends on

• Their geometries
• Their distance of separation
• Number of turns in each coil.
• Nature of medium in the intervening space.

Question 63.
(a) Define the term ‘self-inductance’ and write its S.I. unit.
(b) Obtain the expression for the mutual inductance of two long co-axial solenoids S1 and S2 wound one over the other, each of length L and radii r1 and r2 and n1 and n2 number of turns per unit length, when a current I is set up in the outer solenoid S2. (Delhi 2017)
(a) Self inductance of a coil, equals the magnitude of the efrif induced in it, when the current in the coil, is changing at a unit rate.
S.I. unit : henry
(b) Consider two long solenoids S1 and S2 of same length l such that solenoid S2 surrounds S1 completely.
Let a current I2 flow through S2. This sets up a magnetic flux ϕ1 through each turn of the coil S1

Total flux linked with S1, N1ϕ1 = M12I2… (i)
Where M12 is the mutual inductance between the two solenoids
Magnetic field due to the current I2 and S2 is µ0n2I2
Therefore, resulting flux linked with S1

Question 64.
Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other. (All India 2017)
(a) Mutual inductance of two coils is equal to the magnetic flux linked with one coil when a unit current is passed in the other coil.

Hence, Mutual inductance is equal to the induced emf set up in one coil when the rate of change of current flowing through the other coil is unity.
S.I. unit : Henry, (Weber ampere-1) or (volt second ampere-1)

(b) Consider two long solenoids S1 and S2 of same length l such that solenoid S2 surrounds S1 completely.
Let a current I2 flow through S2. This sets up a magnetic flux ϕ1 through each turn of the coil S1

Total flux linked with S1, N1ϕ1 = M12I2… (i)
Where M12 is the mutual inductance between the two solenoids
Magnetic field due to the current I2 and S2 is µ0n2I2
Therefore, resulting flux linked with S1

Question 65.
Define self-inductance of a coil. Obtain the expression for the energy stored in an inductor L connected across a source of emf. (All India 2017)
(i) Definition of self-inductance :
(a) The self-inductance (L) of a coil equal the magnetic flux linked with it, when a unit current flows through it.
One henry is the self inductance of a coil for which the magnetic flux, linked with it, due to a current of 1A, flowing in it, equals one weber.

(ii) Expression for Magnetic Energy density in an ideal inductor : Instantaneous induced emf in an inductor when current changes through it

Question 66.
State Lenz’s law. Explain, by giving examples that Lenz’s law is a consequence of conservation of energy. (Comptt. Delhi 2017)
Lenz’s law states that “The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.”

As shown in the diagram (a) given, when the north pole of a bar magnet is pushed towards the close coil, the magnetic flux through the coil increases and the current is induced in the coil in such a direction that it opposes the increase in flux. This is possible when the induced current in the coil is in the anticlockwise direction.

Similarly as shown in the diagram (b), just the opposite happens when the north pole is moved away from the coil.

In either case, it is the work done against the force of magnetic repulsion or attraction that gets ‘converted’ into the induced emf.

Question 67.
The current through two inductors of self-inductance 12 mH and 30 mH is increasing with time at the same rate. Draw graphs showing the variation of the
(a) emf induced with the rate of change of current in each inductor
(b) energy stored in each inductor with the current flowing through it.
Compare the energy stored in the coils, if the power dissipated in the coils is the same. (Comptt. All India 2017)

Comparison of energy stored :

Question 68.
The current through two inductors of self-inductance 15 mH and 25 mH is increasing with time at the same rate. Draw graphs showing the variation of the
(a) emf induced with the rate of change of current
(b) energy stored in each inductor with the current flowing through it.
Compare the energy stored in the coils, if the power dissipated in the coils is the same. (Comptt. All India 2017)
Given L1 = 15 mH and L2 = 25 mH

#### Electromagnetic Induction Class 12 Important Questions Long Answer Type

Question 69.
(a) What are eddy currents? Write their two applications.
(b) Figure shows a rectangular conducting loop PQSR in which arm RS of length ‘l’ is movable. The loop is kept in a uniform magnetic field ‘W directed downward perpendicular to the plane of the loop. The arm RS is moved with a uniform speed ‘v’.

Deduce an expression for
(i) the emf induced across the arm ‘RS’,
(ii) the external force required to move the arm, and
(iii) the power dissipated as heat. (All India 2009)
(a) Eddy current : Due to change in magnetic flux, if there is induced current in the volume (bulk) of the material, it is called as eddy currents. It is a necessary evil in an arrangement as it can be used in applications like electric brakes, induction furnaces and dead-beat galvanometers and brings loss of energy with heat production etc.
(b) (i) As the arm RS of length l is moved with a uniform speed, there is a change in area. It is given by dA = Idx = Ivdt

Question 70.
(a) State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy.” Justify this statement.
(b) Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns. (All India 2009)
(a) Lenz’s law (Direction) : “The direction of induced current in a circuit is such as to oppose
the change that causes it. So $$e m f=-\frac{d \phi}{d t}$$
When north pole of a magnet, for example, is brought near a coil, then the nearer end of the coil will acquire north polarity and mechanical work will be needed to cause the relative motion between the coil and the magnet. This work, in fact, converts itself in the form of induced emf. Hence, the phenomenon of electromagnetic induction is in accordance with the law of conservation of energy.

(b) When a current I2 is set up through S2, it in turn sets up a magnetic flux through S1. Let us denote it by ϕ1. The corresponding flux linkage with solenoid S1 is

M12 is called the mutual inductance of solenoid S1 with respect to solenoid S2.
The magnetic field due to the current I2 in S2

We now consider the reverse case :
A current I1, is passed through the solenoid S1 and the flux linkage with coil S2 is,

M21 is called the mutual inductance of the solenoid S2 with respect to solenoid S1

The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoid are very long. Thus, flux linkage with solenoid S2 is

Question 71.
State Faraday’s law of electromagnetic induction. Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The field extends from x = 0 to x = b and is zero for x > b. Assume that only the arm PQ possesses resistance r.
When the arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expressions for the flux and the induced emf. Sketch the variations of these quantities with distance 0 ≤ x ≤ 2b. (All India 2010)

Faraday’s law of electromagnetic induction: “The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit”

Question 72.
(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
(b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(i) Write the expression for the instantaneous value of the emf induced in the wire.
(ii) What is the direction of the emf?
(iii) Which end of the wire is at the higher potential? (All India 2011)
(a) Average power associated with an inductor.
When a.c. is applied to an ideal inductor,current lags behind the voltage in phase by $$\frac{\pi}{2}$$ radian. So we can write the instantaneous values of voltage and current as follows :

Thus the average power dissipated per cycle in an inductor is zero.
(b) (i) The expression for the instantaneous value of the emf induced in the wire is
ε = Blv
(ii) According to Fleming’s right hand rule, “the direction of emf is from west to east”.
(iii) Eastern end of the wire is at the higher electrical potential.

Question 73.
(a) Starting from the expression for the Lorentz magnetic force acting on the free charge carriers of a conductor moving in a perpendicular magnetic field, obtain the expression for the motional emf induced.
(b) Hence deduce the expressions for the power delivered by the source and the power dissipated as Joule heat. (Comptt. Delhi 2012)

Hence, power required to push the conduct is same as the power dissipated.

Question 74.
(a) Describe a simple experiment (or activity) to show that the polarity of emf induced in . a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b) The current flowing through an inductor of self inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) Induced emf versus dI/dt
(iii) Magnetic potential energy stored versus the current. (Delhi 2014)
(a) Lenz’s law : According to Lenz law, “the polarity of the induced emf is such that it opposes a change in magnetic flux responsible for its production.”
Activity :

When the north pole of a bar magnet is pushed towards the coil, the amount of magnetic flux linked with the coil increases. Current is induced in the coil from a direction such that it opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise direction, with respect to an observer.
The magnetic movement ➝ M associated with this induced emf has north polarity towards the north pole of the approaching bar magnet.

Similarly, when the north pole of the bar magnet is moved away from the coil, the magnetic flux linked with the coil decreases. To counter this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.

Question 75.
A metallic rod of length l and resistance R is rotated with a frequency v, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere.
(a) Derive the expression for the induced emf and the current in the rod.
(b) Due to the presence of the current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod.
(c) Hence obtain the expression for the power required to rotate the rod. (Comptt. All India 2012)

The external force required to rotate the rod opposes the Lorentz force acting on the rod/ external force acts in the direction opposite to the Lorentz force.
Power required to rotate the rod

Question 76.
(a) Define mutual inductance and write its S.I. units.
(b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other.
(c) In an experiment, two coils c1 and c2 are placed close to each other. Find out the expression for the emf induced in the coil ct due to a change in the current through the coil c2. (Delhi)
(a) Mutual inductance of two coils is equal to the magnetic flux linked with one coil when a unit current is passed in the other coil.

Hence, Mutual inductance is equal to the induced emf set up in one coil when the rate of change of current flowing through the other coil is unity.
S.I. unit : Henry, (Weber ampere-1) or (volt second ampere-1)

(b) Consider two long solenoids S1 and S2 of same length l such that solenoid S2 surrounds S1 completely.
Let a current I2 flow through S2. This sets up a magnetic flux ϕ1 through each turn of the coil S1

Total flux linked with S1, N1ϕ1 = M12I2… (i)
Where M12 is the mutual inductance between the two solenoids
Magnetic field due to the current I2 and S2 is µ0n2I2
Therefore, resulting flux linked with S1

Question 77.
(a) State Lenz’s law. Use it to predict the polarity of the capacitor in the situation given below :

(b) A jet plane is travelling towards west at a speed of 1800 km/h.
(i) Estimate voltage difference developed between the ends of the wing having a span of 25 m if the earth’s magnetic field at the location has a magnitude of 5 × 10-4 T and dip angle is 30°.
(ii) How will the voltage developed be affected if the jet changes its direction from west to north? (Comptt. All India)
(a) Lenz’s law : It states that “the direction of induced emf is such that it tends to produce a current which opposes the change in magnetic flux producing it.”

Polarity of capacitor: Current induced in the coil will oppose the approach of the magnet, therefore, left face of the coil will act as N-pole and right face S-pole. For this the current in coil will be anticlockwise, as seen from left; and hence the plate ‘A’ of the capacitor will be positive and plate ‘B’ will be negative.

(ii) When the jet changes its direction from west to north, the voltage developed will not be affected.

Question 78.
Define mutual inductance of a pair of coils and write on which factors does it depend.
A square loop of side 20 cm is initially kept 30 cm away from a region of uniform magnetic field of 0.1 T as shown in the figure. It is then moved towards the right with a velocity of 10 cm s-1 till it goes out of the field.

Plot a graph showing the variation of
(i) magnetic flux (ϕ) through the loop with time (t)
(ii) induced emf (ε) in the loop with time t.
(iii) induced current in the loop if it has resistance of 0.1 Ω. (Comptt. All India 2015)
(a) Mutual inductance of two coils is equal to the magnetic flux linked with one coil when a unit current is passed in the other coil.

Hence, Mutual inductance is equal to the induced emf set up in one coil when the rate of change of current flowing through the other coil is unity.
S.I. unit : Henry, (Weber ampere-1) or (volt second ampere-1)

Factors:

• Separation between the coils
• Relative orientation of coils
• Number of turns in the coils

Given: Side of square loop (l) = 20 cm = 20 × 10-2 m

Question 79.
(a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius r1 and the other of radius r2 (r1 < r2) placed coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.
(b) A rectangular coil of area A, having number of turns N is rotated at ‘f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2πf. (All India 2016)
(a) Mutual inductance of two coils is equal to the magnetic flux linked with one coil when a unit current is passed in the other coil.

Hence, Mutual inductance is equal to the induced emf set up in one coil when the rate of change of current flowing through the other coil is unity.
S.I. unit : Henry, (Weber ampere-1) or (volt second ampere-1)

(b) Consider two long solenoids S1 and S2 of same length l such that solenoid S2 surrounds S1 completely.
Let a current I2 flow through S2. This sets up a magnetic flux ϕ1 through each turn of the coil S1

Total flux linked with S1, N1ϕ1 = M12I2… (i)
Where M12 is the mutual inductance between the two solenoids
Magnetic field due to the current I2 and S2 is µ0n2I2
Therefore, resulting flux linked with S1

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = Magnitude of the magnetic field

θ = Angle which normal to the coil makes with field $$\overrightarrow{\mathrm{B}}$$ at any instant t ω = Angular velocity with which coil rotates The magnetic flux linked with the coil at any instant t will be,

Hence Proved.

Question 80.
(a) A metallic rod of length l is moved perpendicular to its length with velocity v in a magnetic field $$\overrightarrow{\mathrm{B}}$$ acting perpendicular to the plane in which rod moves. Derive the expression for the induced emf.
(b) A wheel with 15 metallic spokes each 60 cm long, is rotated at 360 rev/min in a place normal to the horizontal component of earth’s magnetic field. The angle of dip at that place is 60°. If the emf induced between rim of the wheel and the axle is 400 mV, calculate the horizontal component of earth’s magnetic field at the place.
How will the induced emf change, if the number of spokes is increased? (Comptt. All India 2016)
(a) The magnitude of the emf, generated across a length dr of the rod, as it moves at right angles to the magnetic field, is given by
dε = Bvdr

Alternatively, The potential difference across the resistor is equal to the induced emf and equal B x .(rate of change of area of loop), If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ (as shown in the figure) is given by

where [R is the radius of the circle]

(b) Given : Number of spokes (n) = 15,
l = 60 cm = 60 × 10-2 m

• There would be no effect on induced emf, if number of spokes is increased.

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# If Ad is a Median of a Triangle Abc, Then Prove that Triangles Adb and Adc Are Equal in Area. If G is the Mid-point of Median Ad, Prove that Ar (δ Bgc) = 2 Ar (δ Agc). - Mathematics

If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in
area. If G is the mid-point of median AD, prove that ar (Δ BGC) = 2 ar (Δ AGC).

#### Solution

Draw AM ⊥ BC
Since, AD is the median of ΔABC

∴  BD = DC

⇒ BD = AM = DC × AM

⇒  1/2 (BD xx AM ) = 1/2 ( DC xx AM)

⇒ ar (Δ ABC)  = ar (Δ ACD)   ........ (1)

In ΔBGC , GDis the median

∴ ar (BGD)  = area (OGD)      ......... (2)

In ΔACD , CG is the median

∴ area (AGC)  = area (Δ CGD)  ......... (3)

From (1) and  (2) , we have

Area (ΔBGD) = ar (Δ AGC)

But, ar (ΔBGC) = 2ar (BGD)

∴  ar (BGC ) = 2ar (Δ AGC)

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 10 | Page 45


text:
## College Algebra (6th Edition)

Published by Pearson

# Chapter 2 - Functions and Graphs - Concept and Vocabulary Check - Page 223: 4

TRUE.

#### Work Step by Step

$y=x^2-1$ defines $y$ as a function of $x$, as there is only one value of $y$ for each $x$ of the domain.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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OpenGL Advice on pixel collision detection

This topic is 4597 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

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/\ _______
/  \|POLY 2|
/    \--    |
/POLY1|\v|   |
\     |v\|   |
\    |v/|   |
\   |/v|   |
\  /---   |
\/|      |
|______|

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Description too long?

Summary:

Problem 1:
need to load a PNG with transparancy in sdl (or a cross platform method)
need to load that PNG into opengl maintaining transparancy
need to display the png on a quad as a texture with transparancy

Problem 2:
Need to detect which non-transparent pixels overlap between two quads in 2d space with OpenGL.
We can assume the quads are partially or wholly overlapping.

I have a theory on how to do this and I can do the math behind the collisions, I just don't know how to read portions of GLint stored textures and figure out their transparancy on a pixel by pixel basis.

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Not too long, just no easy answer. Just as an example you can use blending. You use GL_SRC_ALPHA for the source to draw the first and GL_DEST_ALPHA to draw the second with GL_ZERO for the destination in both. Within the overlap region the only pixels set at the end are the ones opaque in both and those are set to the values from the second image. You then have to read back the overlap region. That reading back means you would most likely do it faster in the program. Particularly since you are using the frame buffer as a work area for collision detection and have to clear it when you are done. That the images are rotated means doing it in the program isn't a particularly easy task, especially to reproduce to the pixel what you would have gotten from the blending which will match what you see. So personally I can think if many ways to do it, but none I would consider good.

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Quote:
 Original post by LilBudyWizerSo personally I can think if many ways to do it, but none I would consider good.

Yup, that's exactly what I was thinking.

I was thinking that there would possibly be a blending method to somehow visually represent the solution, but I'm an OpenGL noob and wasn't sure if there would be any way to get the "answer" from the visual representation. That lead me to deciding to approach this programatically which would probably still be messy, but at least I'd have all the information (whereas I have no idea how I'd grab the info from the screen with blending especially with backgrounds factored in...)

I have both the textures, I have both their positions and both their rotations, and from there I figured if there was some easy way to read the pixels individually from each texture I could solve the problem from there. I'm not worried about "figuring out" how to manually do the rotations and check the results once I have the textures loaded into an easily checkable format (such as my array where transparant pixels are 0 and opaque are 1), but the performance might not be ideal. My plan is to try implementing before optimizing though. It is just a 2d game afterall and we cut down a lot on the testing by only performing the per-pixel manual test on already colliding quads.

*edit: PS, thank you for your words. I'm definitly + ranking anyone with help.

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Does it have to be a pixel-based solution? Or would a geometric solution be an option? If you're interested in other options, here are a couple of questions:

1. What kinds of shapes are you dealing with for your sprites? Are they convex shapes (like a circle or box) or concave/arbitrary (like an asteroid, spaceship, or character)?

2. What do you intend to do with the collision information? Do you want to resolve the collision (i.e. move the objects by the shortest vector that will resolve the interpenetration), or just blow them up (like in an asteroids game)?

If you're still looking for options, the answers to those questions will help in making reasonable suggestions.

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Another option would be to store polygon information about the shape of the limbs. But I'd have to do this for each different texture which I want to avoid.

What I have being set up:
-Skeleton system with joints and bones.
-Bones are represented by textured quads.
-the textures on the quads (and the width and length of those quads) can be swapped as the skeleton shifts positions based on desired keyframe positions. (Ex: player looks towards us, then turns which squashes the quad's width (width is the perpendicular scale to length where length of the quad is equal to the length of the bone) the quad representing the torso and flips to a different texture)
-attacks are to be precise as the character's arm has a lot of control and it's the combat system which will be the main innovation.

I'm toying with the idea of damage as well. Damage could be indicated by detachable limbs if the joint is smacked hard enough (or an area around the joint rather) and could be indicated by blood as well... The difficulty with blood splatters is the high dynamic nature of the characters and the fact that the textures change to reflect position. Example. character faces us, then turns and faces away from the camera which should obscure the blood. How to represent this? I'll work on that, perhapse storing actual pseudo 3d rotation of the blood splatters as well and factoring those in somehow... But that's just an idea and I'll look into it after I get collisions.

Detachable limbs is a definite and per-pixel collisions is important to me because it can also be used to figure out how much damage is done. Sword intercepts with 30 pixels does more damage than 10 pixels. Modify that by the body part attacked (ie: arms do less damage than head or torso) and the damage system could be fairly good too.

The shape of the actual textures on the quads is arbitrary and may include concave shapes (picture spikes on armour which hook up and back, or fingers) however ultimately I need to know how to read the textures pixel by pixel and that's all I need to know, it's not the same as creating a polygon and detecting the collision unless you derive a polygon from the textures and then run an operation, but since the textures exist in a limited amount of pixels I think it would be more efficient to just compare pixels and be done with it. I was just looking for someone who might have done some similar pixel by pixel reading from the opengl texture format before, and hopefully finding out the included alpha value of those individual pixels if there is one on opengl textures.

In essence, assume I know that I want pixel based collisions as I am dealing with textures, not polys. I've thought about approximations with the polys being stored along with each texture, but that will take more setup for each character and will not be as accurate (though it might be faster.)

I'm pretty sure I want what I've described. I'm not 100% against other options or somehow detecting the edges of the textures, that's fine too, but I don't think it will be easier than just reading the pixels for the portions that are intercepting.

I am not asking about collision detection primarily, that's just what I'm accomplishing with this.

I'm asking about how to iterate through an opengl texture's individual pixels and how to determine the alpha value if that's stored or not.

*edit: has anyone ever drawn on a texture with the mouse or a tablet? Like a painting program? I'd assume you'd need to know how to do something like this for that to work.

[Edited by - M2tM on December 16, 2005 6:30:54 PM]

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Well, you use glGetTexImage to read the pixels of a texture. You specify a format and data type just as you did when you loaded the texture. What the original format was or how it is stored internally doesn't matter. It will be returned in the format you specified.

One thought is don't rotate your quads. Rather scale your quads so they can completely contain the rotated texture. Give the textures a transparent border color and clamp them. Then use the texture coordinates to actually rotate the textures. Then you always draw screen aligned quads. All you have to do is read the overlapped regions back.

If multiple sprites may collide it would most likely be best to draw all of them before you start reading back the overlapped regions. That might help the concurrency between the GPU and the CPU, i.e. it can be drawing quads while you are calculating quads. You're going to sync on the read though so you might as well do all your reads at once so it is logically one sync point.

With a bit of planning you might be able to avoid a clear and complete redraw. Draw all your sprites, deal with the overlap and just redraw those. I think you can use the alpha test to draw the background last. I believe the transparent pixels set the depth even though they are not visible so the depth test would leave blank the transparent pixels.

It most likely wouldn't make a differance with a middle of the road 3D graphics accelerator, but it might make a big differance with a 2D graphics card. Using quads and textures may mean you don't use the blitting hardware at all which would make your fillrate downright brutal. If so then you don't want to redraw anything you don't have to.

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# Neither thick set nor syndetic set but piecewise syndetic set

I am finding an example, which is neither thick set nor syndetic set but piecewise syndetic set. After too many many trials I got nothing. Can you help me to find this example? Thank you.

• Please define "example" (of what?), "thick" and the unusual word "syndetic". – Dr. Lutz Lehmann Dec 2 at 9:22


text:
# Best basis for taxation

by DaleSpam
Tags: basis, taxation
HW Helper
P: 3,684
 Quote by baywax The table would have to include every industry and every department of government to give a good representation of the country. When each party can see the effects of their policies on the other a better understanding of the whole can be achieved and policies can shift accordingly. But I know its much more complicated than that. For example, when the province of BC looks good because we have timber to sell, people forget about the soft wood lumber tariff imposed by the US on that commodity and they forget about the devastation caused by the pine beetle which has wiped out the soft wood industry in BC. We have what are called have and have not provinces. BC looks like a have but in actual fact the have we have is not. So many factors get lost in a 3-5 day meeting.
But representing every industry will result in a large enough meeting that some will be ignored... taking us back to the starting point. And this doesn't even deal with the issue of things that are beneficial only to companies but not consumers -- legalizing collusion, for example.
Mentor
P: 15,058
 Quote by CRGreathouse The widget companies' lower tax liability is not a benefit -- the only possible benefit is the government's tax revenue (which may increase or decrease).
You must have missed the rolls-eyes smilely at the end of the paragraph. To make it very explicit: Any statements regarding the benefits accrued by the government forcing companies to go out of business were strictly tongue-in-cheek.
HW Helper
P: 3,684
 Quote by baywax If the government used the FIELDS formulae where you charge very little on a lot of products... thus huge volume creates huge income... then there would be an even spread of the burden.
In what way should the burden be spread around?

I guess I'd like to split the problem of taxation into two parts: ethical and positive/non-normative. First we decide what would be fair to tax, then we decide (based on what fairness principles are accepted) what method of taxation would be least harmful.
PF Gold
P: 2,215
 Quote by CRGreathouse In what way should the burden be spread around? I guess I'd like to split the problem of taxation into two parts: ethical and positive/non-normative. First we decide what would be fair to tax, then we decide (based on what fairness principles are accepted) what method of taxation would be least harmful.
Going back to the General Services Tax. This applied to every purchase of most goods with regulations stating things like buying a dozen buns or a loaf of bread or a jug of milk would be exempt from the tax... but the companies creating these items collects GST on most items... passing the Government's taxation along to the consumer in full view of the consumer. In fact each service company charges GST for the govt and hands it in quarterly. The cut to begin with was 7%. Now its been lowered for popularity issues to 5%... just as this economic throttling has taken place. This taxation technique helped take the burden of running a nation off the big producers to some degree. It was like the FIELDS model in that it applied across the board to every service and many items. Often only representing a few cents here and there for the low-middle income consumer.

Luxury tax is another item that works to a degree. It hands the responsibility of a person's choice back to them in the form of a tax that will pay for search and rescue, airports, polluting manufacturing clean up costs, smoother roads etc.... because the car or stereo or Cessna manufacturer can't build that tax into there pricing but they can blame the government for it. In fact the rebates of up to $2000 on gas efficient vehicles provide the manufacturer with incentives and a sign that the government gives back.. sometimes. Sci Advisor HW Helper P: 3,684  Quote by baywax Luxury tax is another item that works to a degree. It hands the responsibility of a person's choice back to them in the form of a tax that will pay for search and rescue, airports, polluting manufacturing clean up costs, smoother roads etc. The purpose, presumably, of a luxury tax is to tax the rich more than the poor, especially when they're spending on 'frivolous' things instead of investing in the stock market or paying for their kids' college tuition. My point about separating ethics from non-normative economic analysis applies here: I'd like to compare the efficiency of • the current system • the current system with higher lux taxes and lower income tax on the rich • the current system with lower lux taxes and higher income tax on the rich  Quote by baywax the car or stereo or Cessna manufacturer can't build that tax into there pricing but they can blame the government for it. Car, stereo, and Cessna manufacturers do build the taxes into their prices. If car companies no longer needed to pay taxes, they could make more money by dropping their price (not by the full amount of the tax, but partway) in order to sell more units. PF Gold P: 2,215  Quote by CRGreathouse Car, stereo, and Cessna manufacturers do build the taxes into their prices. Sorry... what I was saying there was that the GST or Luxury Tax is alway added to the price of a luxury item... it shows up on the invoice as a government imposed tax... and is not hidden... unethically.... from the consumer. Sci Advisor HW Helper P: 3,684  Quote by baywax Sorry... what I was saying there was that the GST or Luxury Tax is alway added to the price of a luxury item... it shows up on the invoice as a government imposed tax... and is not hidden... unethically.... from the consumer. Ah. I'm not talking about that, though; I'm talking about how prices really change. If a company makes luxury watches which sell for$500 (making a profit of $120 per watch) and suddenly there is a luxury tax of 10% on the watches, I expect the company to lower prices (perhaps to$480, lowering profit per watch to $100) to keep buyers from switching to alternatives that aren't taxed. This is because, in essence, the company can make the watch$22 cheaper by losing only $20 per watch, so there's more benefit from the 'increased sales' (relative to not dropping the prices post-tax) than from the higher margin. Sci Advisor P: 2,283  Quote by D H ...(2) the rich do benefit more from government spending than do the poor. While the rich may not drive much more than the middle class, the companies whose stocks they own do benefit from improved roadways. Making it easier for people to get to work makes it easier form companies to make a profit. The rich benefit from government services to people other than themselves. IMO, I don't think the rich benefit from government services more than anyone else. In fact, using your truck example, passenger vehicles account for 93% of highway usage compared to the 7% of commercial vehicles. Plus commercial vehicles have higher associated fees than passenger vehicles, and the trucking company is providing a job for the driver (who isn't rich). CS Sci Advisor P: 2,283  Quote by DaleSpam Also, the rich directly benefit more because they have more assets that are protected by the police and military. As a Libertarian I disagree that John should pay for a benefit to Fred, regardless of the fact that John, as Fred's employer, gets some indirect benefit, but I think even so that there can be a direct justification for the rich to pay more than the poor. The military (and police) protect the US as a whole, regardless of how much each individual person makes. I don't see a benefit for the rich here either. I mean, we all live in the same place that needs protecting don't we (i.e. the US)? CS Mentor P: 16,945  Quote by stewartcs The military (and police) protect the US as a whole, regardless of how much each individual person makes. I don't see a benefit for the rich here either. I mean, we all live in the same place that needs protecting don't we (i.e. the US)? Yeah, I see your point. I guess the idea is simply that the rich have a bigger "slice" of the "US as a whole"-pie. So they can be considered to derive more benefit from the same protection. At the same time, that really brings up the issue of income vs. wealth, how do you best measure the slice? Sci Advisor P: 2,283  Quote by DaleSpam I guess the idea is simply that the rich have a bigger "slice" of the "US as a whole"-pie. So they can be considered to derive more benefit from the same protection. At the same time, that really brings up the issue of income vs. wealth, how do you best measure the slice? Do they really have a bigger slice of the pie though as a whole? There are by far less "rich" people than "poor" people in the US. I think they make up something like 1% of the people in the US which would be a rather small slice. CS Sci Advisor HW Helper P: 3,684  Quote by DaleSpam Yeah, I see your point. I guess the idea is simply that the rich have a bigger "slice" of the "US as a whole"-pie. So they can be considered to derive more benefit from the same protection. At the same time, that really brings up the issue of income vs. wealth, how do you best measure the slice? I think this is a valid point -- the rich do get more benefit. But assuming that the police/military protection applies to both the person and the property, and that both are of positive value (we prefer living to being killed, we prefer having our house to having someone burn it down), even a straight wealth tax (the property tax is a close relative) is 'too progressive', and that the tiered income tax structure is right out. Sci Advisor HW Helper P: 3,684  Quote by stewartcs Do they really have a bigger slice of the pie though as a whole? There are by far less "rich" people than "poor" people in the US. I think they make up something like 1% of the people in the US which would be a rather small slice. I don't know if that's relevant. DaleSpam was just suggesting that each rich person gets more benefit from government protection than each poor person, which seems reasonable to me. The government doesn't protect my vacation home because I don't have one. But obviously how many rich people there are depends on your definition of rich. By Victorian standards I'd imagine the majority of people in the US are rich. Sci Advisor HW Helper P: 3,684  Quote by stewartcs IMO, I don't think the rich benefit from government services more than anyone else. In fact, using your truck example, passenger vehicles account for 93% of highway usage compared to the 7% of commercial vehicles. Plus commercial vehicles have higher associated fees than passenger vehicles, and the trucking company is providing a job for the driver (who isn't rich). I don't know if they benefit a lot more or just a little more -- people have argued both on this thread -- but I do think that they (sensibly) benefit more. But I'm not saying that the total benefit to all rich people is greater than the total benefit to all nonrich people, just that the average rich person gets more gross benefit from the government than the average nonrich person does -- for some reasonable value of 'rich', anyway. Sci Advisor P: 2,283  Quote by CRGreathouse I don't know if that's relevant. DaleSpam was just suggesting that each rich person gets more benefit from government protection than each poor person, which seems reasonable to me. The government doesn't protect my vacation home because I don't have one. The government still protects the adjacent/surrounding homes that probably are not vacation homes. It's not like the "rich" people use the government as private security for their vacation homes or whatever other property they own. The security/police/military is already in place and would exist regardless of whether or not the "rich" had the extra home or other property. So I don't see how they benefit more.  Quote by CRGreathouse But obviously how many rich people there are depends on your definition of rich. By Victorian standards I'd imagine the majority of people in the US are rich. I agree the definition of rich is quite subjective. Admittedly I'm not familiar with what Victorian standards were like, I imagine if one were to adjust the value of money for inflation, etc. then it wouldn't be too much of a difference in the terms. CS Sci Advisor P: 2,283  Quote by CRGreathouse ...just that the average rich person gets more gross benefit from the government than the average nonrich person does -- for some reasonable value of 'rich', anyway. Just out of curiosity, how did you form that opinion? I seem to hear this statement a lot but I've never really heard a reasonable explanation as to why other than that's just the way it is. CS Mentor P: 15,058  Quote by stewartcs The government still protects the adjacent/surrounding homes that probably are not vacation homes. It's not like the "rich" people use the government as private security for their vacation homes or whatever other property they own. The security/police/military is already in place and would exist regardless of whether or not the "rich" had the extra home or other property. So I don't see how they benefit more. On that basis, I hereby petition my local government to stop levying property taxes on me (at least the portion that goes to the police department) because the security/police is already in place and would exist regardless of whether or not I personally pay my share of that protection. Sci Advisor HW Helper P: 3,684  Quote by stewartcs I agree the definition of rich is quite subjective. Admittedly I'm not familiar with what Victorian standards were like, I imagine if one were to adjust the value of money for inflation, etc. then it wouldn't be too much of a difference in the terms. Well, for starters, the Victorians didn't have computers, cars, air conditioning, vaccines, or well-insulated houses. In modern America a person can own a car and live in an insulated apartment with heat and air conditioning while still falling below the poverty line.  Quote by stewartcs The government still protects the adjacent/surrounding homes that probably are not vacation homes. It's not like the "rich" people use the government as private security for their vacation homes or whatever other property they own. The security/police/military is already in place and would exist regardless of whether or not the "rich" had the extra home or other property. So I don't see how they benefit more. I agree that the security would exist without those homes, but the rich still get that benefit (even if it's totally unintentional).  Quote by stewartcs Just out of curiosity, how did you form that opinion? I seem to hear this statement a lot but I've never really heard a reasonable explanation as to why other than that's just the way it is. First of all, let me state that this is not a bad state of affairs. If a nonrich person pays$8,000 in taxes and gets $12,000 in government benefits while a rich person pays$100,000 in taxes and gets $10,00 in benefits, that's a pretty unfair situation. Saying that the rich person in this example gets more than$12,000 in benefits (perhaps only $13,000) doesn't seem like much to ask. I'm not saying that they get more proportional benefit -- in fact I feel that this is false. • Police protection is determined in part by location: better areas get better police protection because local policing is paid for by local taxes. Rich people live in better areas. • Rich people have more to lose. If the government shuts down and a mob destroys my house, I'm out$**,000. If a mob destroys a rich person's mansion, they're out \$*,*00,000.
• Rich people get indirect benefits. If you own stock in any company that uses US infrastructure (that would be pretty much every US company) you benefit from that company's gain.

 Related Discussions Calculus & Beyond Homework 2 Quantum Physics 0 Advanced Physics Homework 0 Calculus & Beyond Homework 7 Differential Geometry 37


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## General Chemistry 10th Edition

We know that when a precipitate fails to form by adding HCl to the solution,then this shows that no silver ion is present. Similarly, when a precipitate fails to from by adding $H_2S$ shows that no copper(II) ion is present.


text:
# Making a CGridView row clickable.

Last updated:

This is how you would make each row of you CGridView a link to the item displayed in it. The most important configurations is selectionChanged.

<?php
$this->widget('zii.widgets.grid.CGridView',array( 'id'=>'items-grid', 'dataProvider'=>$items,
'selectionChanged'=>"function(id){window.location='"
.Yii::app()->urlManager->createUrl('item/view',array('id'=>''))."' +
\$.fn.yiiGridView.getSelection(id);}",
'columns'=>array(
//more stuff.....


text:
# How can the solid state structure of cyclopentadienyllithium be explained?

In the solid state, LiCp adopts a polymeric multidecker structure where Li atoms are sandwiched between two Cp rings (shown in this diagram from Organometallics 1997, 16 (17), 3855–3858):

If we consider the η5-binding mode of the Cp rings, then it seems that the number of electrons around Li is more than eight. However, Li can't extend its octet. How can the structure and bonding be explained?

• It is ionic bonding not covalent bonding. A lithium cation is between two cyclopentadienyl anions in the 3D structure of the solid, but the chemical formula is 1:1. – MaxW Sep 12 '18 at 15:17
• I think the question may well be more interesting than "it's ionic bonding". MeLi and other alkyllithiums, for example, can hardly be described as purely ionic - the structure and bonding is quite complicated. I wouldn't be surprised if there was significant covalency in CpLi, too, although I might of course be wrong, and a cursory flip through my usual bunch of references doesn't turn up any more useful info. – orthocresol Sep 12 '18 at 15:36
• Well, as you can easily count it's six electrons per 1 Li as all rings are bound to two Li atoms. This type of boding is present in multidecker sandwich complexes. – Mithoron Sep 12 '18 at 16:01
• @Mithoron...I understand your point but in that case, its octet is not fulfilled, then how this structure is so stable? why the molecule does not adapt any more complicated structures? – Deepta Chattapadhyay Sep 12 '18 at 16:27
• It didn't ping me (because of dots presumably). Li makes electron deficient molecules almost all the time. It's middle ground between, say, Rb with close to zero valence electrons and carbon with octet most of time. – Mithoron Sep 12 '18 at 19:08

## 1 Answer

This summary identifies several calculations indicating that covalent interactions between lithium and cyclopentadienyl moieties do occur; the interaction is mostly but not purely ionic.

Draw the occupied $\pi$ orbitals of the cyclopentadienyl groups and, for comparison, draw the $s, p_x, p_y, p_z$ orbitals on lithium with the $z$ axis along the length of the chain. Observe:

• The lithium $s$ and $p_z$ orbitals overlap the lowest energy cyclopentadienyl $\pi$ orbitals.

• The lithium $p_x$ and $p_y$ orbitals overlap the bonding degenerate pair on the cyclopentadienyl ring.

With these overlaps, we can construct three lithium-cyclopentadienyl bonding combinations per cyclopendienyl group, one corresponding to each cyclopentadienyl $\pi$ bonding orbital. We then occupy these orbitals with six electrons per ring. In effect we have extended the aromatic coupling of the cyclopentadienyl ion into the third dimension with the lithium orbitals.

Heavier alkali metals do not work as well because their orbitals are too diffuse, but they could overlap larger rings.


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Strange Attractors for Periodically Forced Parabolic Equations
Kening Lu, Brigham Young University, Provo, UT, Qiudong Wang, University of Arizona, Tucson, AZ, and Lai-Sang Young, Courant Institute of Mathematical Sciences, New York University, NY
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Memoirs of the American Mathematical Society
2013; 85 pp; softcover
Volume: 224
ISBN-10: 0-8218-8484-0
ISBN-13: 978-0-8218-8484-3
List Price: US$69 Member Price: US$55.20
Order Code: MEMO/224/1054

The authors prove that in systems undergoing Hopf bifurcations, the effects of periodic forcing can be amplified by the shearing in the system to create sustained chaotic behavior. Specifically, strange attractors with SRB measures are shown to exist. The analysis is carried out for infinite dimensional systems, and the results are applicable to partial differential equations. Application of the general results to a concrete equation, namely the Brusselator, is given.

• Introduction
• Basic definitions and facts
• Statement of theorems
• Invariant manifolds
• Canonical form of equations around the limit cycle
• Preliminary estimates on solutions of the unforced equation
• Time-$$T$$ Map of forced equation and derived $$2$$-D system
• Strange attractors with SRB measures
• Application: The Brusselator
• Appendix A. Proofs of Propositions 3.1-3.3
• Appendix B. Proof of Proposition 7.5
• Appendix C. Proofs of Proposition 8.1 and Lemma 8.2
• Bibliography