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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Saturday, March 26, 2011

### AP Physics B & C - Multiple Choice Practice Questions Involving Electromagnetic Induction

“You must not lose faith in humanity. Humanity is like an ocean; if a few drops of the ocean are dirty, the ocean does not become dirty.”

– Mahatma Gandhi

Questions involving electromagnetic induction have been discussed on several occasions on this site. You can access all of them either by trying a search for ‘electromagnetic induction’ using the search box provided on this page or by clicking on the label ‘electromagnetic induction’ below this post. Questions on electromagnetic induction are generally interesting and so the aspirants of AP Physics B as well as AP Physics C will definitely ‘enjoy’ them. The following questions are meant for gauging the depth of your knowledge and understanding in this area:

(1) A particle of mass m and charge –q moves with uniform speed from point A to point B along a straight path AOB in the positive y-direction. A plane conducting circular loop is arranged near the path AOB (as shown in the figure) so that the loop and the path AOB are in the same plane. Pick out the correct statement regarding the current induced in the loop during the interval the charged particle moves from A to B:

(a) The induced current increases and flows clockwise in the loop

(b) The induced current increases and flows anticlockwise in the loop

(c) The induced current increases, reaches a maximum and then decreases; but it flows clockwise in the loop

(d) The induced current increases, reaches a maximum and then decreases; but it flows anticlockwise in the loop

(e) The induced current changes its direction

This question may confuse many among you quite a lot. The particle has a negative charge and that too may enhance your doubts.

Think of the moving charge as a moving current element. Since the charge is negative and is moving along the positive y-direction, the direction of the conventional current is along the negative y-direction. At the circular loop this current element gives rise to a magnetic field perpendicular to the plane of the loop, directed towards the reader. When the charged particle is at A, the magnetic flux through the loop is small and when it is nearest to the circular loop (at O), the magnetic flux through the loop is maximum. The increasing magnetic flux induces a current in the loop. By Lenz’s law the induced current must flow in the clockwise sense (so that the magnetic flux produced by the induced current opposes the increasing magnetic flux produced by the moving charge).

When the charge moves from O to B, the magnetic flux linked with the loop decreases. The current induced in the loop now changes its direction since the magnetic flux produced by the induced current has to oppose the decreasing magnetic flux produced by the moving charge. The correct option is (e).

[The last option can be further refined and stated as follows:

(e) The induced current flows in the clockwise sense and anticlockwise sense respectively when the charged particle traces the paths AO and OB].

(2) A rectangular conducting loop PQRS (Fig.) is located in a magnetic field B which acts perpendicularly into the plane of the loop. The magnetic field is constant and it exists over a wide region, well beyond the extent of the loop. The loop is moved, within the magnetic field, with uniform velocity v in a direction mutually perpendicular to the direction of the magnetic field and the side QR. Pick out the correct statement regarding the induced current (if any):

(a) The induced current in the side PQ flows from P to Q.

(b) The induced current in the side RS flows from S to R.

(c) There is no induced current in the loop since the voltage induced in the side PS as well as QR is zero.

(d) There is no induced current in the loop since no voltage is induced in any of the sides of the loop.

(e) There is no induced current in the loop since the voltages induced in the sides PS and QR are equal and they act in opposition in the loop.

No current is induced in the loop since the magnetic flux linked with the loop is the same in all positions of the loop. So there is no change of flux to produce an induced emf in the loop. Even though the last three options indicate that the induced current in the loop is zero, the reason for this is correct in the case of option (e) only. Equal motional emfs are generated in the sides PS and QR and these act in the loop as two voltage sources in opposition.

(3) A copper rod moves at a constant velocity in a direction perpendicular to its length. A constant magnetic field mutually perpendicular to the rod and its direction of motion exists in the region. Pick out the correct statement from the following:

(a) The electric potential at the centre of the rod is maximum when the rod moves.

(b) The electric potential at the centre of the rod is minimum when the rod moves.

(c) The electric potential is the same at all points of the rod when it moves.

(d) An electric field is produced in the rod when it moves.

(e) The electric potential is zero at the centre of the rod and increases towards the ends

When the rod moves at right angles to the magnetic field, the free electrons in the rod experience magnetic (Lorentz) force and shift towards one end. Thus there is an accumulation of negative charges at one end of the rod and consequently there exists an electric field in the rod [Option (d)].

The following question is specifically for AP Physics C aspirants:

(4) The series combination of an inductance L and a resistance R shown in the adjoining figure is part of an electric circuit which drives a current i through the combination. When the current through the combination is 6 mA and is decreasing at the rate of 10 mA s–1, the potential difference between points A and B is 6 mV. When the current through the combination is 6 mA and is increasing at the rate of 10 mA s–1, the potential difference between points A and B is 12 mV. The values of the inductance L and resistance R are respectively

(a) 4 H and 6 Ω.

(b) 3 H and 6 Ω

(c) 0.5 H and 3.5 Ω

(d) 0.3 H and 1.5 Ω

(e) 1.5 H and 1.5 Ω

It is the potential difference between the points A and B that drives a current through L and R. When the current in the circuit decreases, the inductor has to try to oppose the decrease by inducing a voltage in it to aid the 6 mV across the points A and B. Therefore Kirchhoff’s voltage equation for this case is

– L(di/dt) + iR = P.D. between A and B

[We could have better written this as (P.D. between A and B) + L(di/dt) = iR]

Substituting known values, we have

– L×10×10–3 + 6×10–3 R = 6×10–3

Or, – L×10 + 6R = 6 -----------(i)

When the current in the circuit increases, the inductor has to try to oppose the increase by inducing a voltage in it to oppose the 12 mV across the points A and B. Therefore Kirchhoff’s voltage equation for this case is

L(di/dt) + iR = P.D. between A and B

[We could have better written this as (P.D. between A and B) – L(di/dt) = iR]

Substituting known values, we have

L×10 + 6R = 12 -----------(ii)

Adding Eq(i) and Eq(ii) we have 12 R = 18 from which R = 1.5 Ω.

On substituting the value of R in Eq(ii) we obtain L = 0.3 H [Option (d)].

## Sunday, March 13, 2011

### AP Physics C – Answer to Free Response Practice Question on Electrostatic Potential Energy

“Genius is one per cent inspiration and ninety nine per cent perspiration.”

Thomas A. Edison

A free response practice question on Electrostatic Potential Energy was given to you in the post dated 11th March 2011. As promised, I give below a model answer along with the question:

A thin circular ring of radius R has positive charges sprayed uniformly along it so that the linear charge density along it is λ. The ring is located in the y-z plane with its centre at the origin of a right handed Cartesian co-ordinate system, in a space lab, where the effect of gravity is negligible. From the point P [–(√15)R, 0, 0] shown in the figure, a particle of mass m and charge +q is projected with speed v along the positive x-direction. Now answer the following:

(a) Which one of the following statements (i), (ii) and (iii) regarding the electric field vector at P (due to the charges on the ring) is correct? [Put a tick (√) mark against your choice].

(i) Electric field at P is along the positive x-direction ……..

(ii) Electric field at P is along the negative x-direction ……..

(iii) Electric field at P is zero ……..

(b) Determine the electrostatic potential energy of the charge q when it is located at P.

(c) Determine the electrostatic potential energy of the charge q when it is located.at the centre of the ring.

(d) Determine the minimum value of the speed of projection for the charged particle to pass through the ring.

(e) Explain what happens to the charged particle if the speed of projection is less than the minimum value mentioned in part (d) above.

(a) (ii) Electric field at P is along the negative x-direction …√…..

Since the point P is on the axis of the charged ring, the electric field at P is directed along the axis. Since the charge on the ring is positive, the electric field is directed from the ring to the point P. So the field is along the negative x-direction.

The direction of the field due to the charge at a point A on the ring is directed along the line AP. This field can be resolved into rectangular components, one along the axis of the ring and the other normal to the axis. When we consider the electric field due to an equal charge situated at the diametrically opposite point B on the ring, we find that the normal components of the fields due to A and B are equal and opposite so that they get canceled. But the axial components are in the same direction (negative x-direction) and they get added up. This is why the field at P is along the negative x-direction.

(b) The total charge Q on the ring is given by

Q = 2πR λ.

The distance r of the point P from the charges is given by

r = [R2 + 15 R2]1/2 = 4R

Therefore, the electric potential V1 at the point P is given by

V1 = (1/4πε0)(Q/r) where ε0 is the permittivity of free space.

Substituting for Q and r, we have

V1 = (1/4πε0)( 2πRλ /4R) = λ/8ε0

The electrostatic potential energy E1 of the charge q when it is located at P is given by

E1 = V1q = λq/8ε0

(c) The electric potential V2 at the centre of the ring is given by

V2 = (1/4πε0)(Q/R) = (1/4πε0)( 2πRλ /R) = λ/8ε0 = λ/2ε0

The electrostatic potential energy E2 of the charge q when it is located at the centre of the ring is given by

E1 = V2q = λq/2ε0

(d) When the particle of mass m and charge +q is projected (along the axis) towards the ring, its potential energy increases until it reaches the centre of the ring. The increase in potential energy is at the cost of the kinetic energy of the particle. Once the particle just passes the centre of the ring, it can continue to move forward and gain kinetic energy at the cost of its potential energy.

If the initial kinetic energy with which the particle is projected from the point P is equal to the difference between its electrostatic potential energies at the points P and the centre of the ring, the particle will reach the centre of the ring with zero speed. Therefore, the minimum kinetic energy with which the charged particle is to be projected so as to pass through the ring is equal to the difference between its electrostatic potential energies at the points P and the centre of the ring. Therefore, we have

½ mv2 = E1 E2 where v is the minimum speed with which the charged particle is to be projected.

Substituting for E1 and E2 we have

½ mv2 = (λq/2ε0) – (λq/8ε0) = 3λq/8ε0

This gives v = (3λq/4ε0m)1/2

(e) If the speed of projection of the charged particle is less than the minimum value mentioned in part (d) above, it will move towards the ring with its speed decreasing gradually. Its kinetic energy will become zero before reaching the centre of the ring and it will momentarily come to rest. Under the action of the electric field directed along the negative x-direction, the particle will turn back and will retrace its path and will move past the point P all the way to infinite distance from the ring before coming to rest.

## Friday, March 11, 2011

### AP Physics C - Free Response Practice Question on Electrostatic Potential Energy

“I believe in standardizing automobiles, not human beings.”

Albert Einstein

Today I’ll give you a free response practice question involving electrostatic potential energy. You may answer it yourself so that you will realize your strength and weakness in answering free response questions within the stipulated time. Of course, I’ll give you a model answer later, as usual.

Here is the question:

A thin circular ring of radius R has positive charges sprayed uniformly along it so that the linear charge density along it is λ. The ring is located in the y-z plane with its centre at the origin of a right handed Cartesian co-ordinate system, in a space lab, where the effect of gravity is negligible. From the point P [–(√15)R, 0, 0] shown in the figure, a particle of mass m and charge +q is projected with speed v along the positive x-direction. Now answer the following:

(a) Which one of the following statements (i), (ii) and (iii) regarding the electric field vector at P (due to the charges on the ring) is correct? [Put a tick (√) mark against your choice].

(i) Electric field at P is along the positive x-direction ……..

(ii) Electric field at P is along the negative x-direction ……..

(iii) Electric field at P is zero ……..

(b) Determine the electrostatic potential energy of the charge q when it is located at P.

(c) Determine the electrostatic potential energy of the charge q when it is located.at the centre of the ring.

(d) Determine the minimum value of the speed of projection for the charged particle to pass through the ring.

(e) Explain what happens to the charged particle if the speed of projection is less than the minimum value mentioned in part (d) above.

The above question is worth 15 points and you may take about 15 minutes for answering it. Try to answer it. I’ll be back soon with a model answer for your benefit.