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)].

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 11^th 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 ……..

Justify your answer.

(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.

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 ……..

Justify your answer.

(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.

Free Body Diagrams for AP Physics B and C

“To give pleasure to a single heart by a single act is better than a thousand heads bowing in prayer”

– Mahatma Gandhi

As AP Physics B and C aspirants you need to master the technique of drawing free body diagrams (FBD) without mistakes. You might have encountered many situations, especially in mechanics, where you were required to draw correct free body diagrams. Drawing free body diagrams at the level expected of you is ordinarily simple. Yet there may be many among you who have genuine doubts in this area.

A free body diagram of an object is one in which all the forces acting on the object are pictorially represented. The object may be the part of a complex system. But in the free body diagram of an object, you need to show the object alone (represented by a simple geometrical figure such as a dot or a small square or a small rectangle or the outline of the object) along with the forces represented by directed line segments (arrows). The length of the arrows, as usual in vector representation, should represent the magnitude of the force while the direction of the arrow should indicate the direction of the force. Since the free body diagram is a pictorial representation of the object along with the forces acting on it, used for simplifying your analysis of a problem, you need not worry about the exact length of the arrow and its exact direction. But the labeling of the magnitude and direction of the known forces should be correct.

In AP Physics free response questions often you may be asked to represent an object (such as a skier moving along a ramp or a body floating on the surface of a liquid) by a dot and to draw and label the forces that act on the object. Understand that you are being prompted to draw a simple free body diagram.

We will now draw the free body diagrams of some typical objects and systems in typical situations so that you will become more confident for facing your AP Physics Exam.

(1) A box of mass M at rest on a horizontal table:

The box can be represented by a small rectangle in the free body diagram as shown in FBD (i) in the figure. The forces acting on the box are the gravitational pull (weight) F_Gr and the normal force F_Normal exerted by the surface of the table_. These are of equal magnitude, but opposite in direction (F_Gr being downwards and F_Normal upwards. Surely, F_Gr = Mg.

[Even if you dispense with the symbol F_Gr for the gravitational pull and just write the weight Mg as label for it, your free body diagram will be alright].

If you were asked to represent the box by a dot, you would draw the free body diagram as shown in FBD (ii).

You know that the weight of the box acts through the centre of gravity of the box where as the normal force is exerted at the contact surface. You can draw the force vectors taking these into consideration; but there is no harm in showing them as in FBD (i) since the body has neither translational motion nor rotational motion since the equal and opposite forces have their lines of action coinciding.

When you are asked to represent the object by a dot, the question setter is in fact making things simpler for you.

(2) A block of mass M at rest on an incline:

In this case three forces act on the object. They are:

(i) The weight Mg of the block

(ii) The normal force F_Normal exerted by the incline and

(iii) The force of friction F_fr preventing the block from sliding down

The object is represented by a small rectangle in the free body diagram shown in FBD (i) in the figure. The frictional force F_fr acts along the surface of the block while the weight of the block acts through the centre of gravity of the block. The normal force in this case is opposite to the component of the weight normal to the inclined surface. Since the block does not rotate, the three forces acting on it have to be concurrent and so you need not be confused as to how to show the lines of action of the three forces. If you are asked to represent the block by a dot as shown in FBD (ii) you will be relieved of confusion.

[Note that the component of weight normal to the inclined surface is Mg cosθ where θ is the inclination of the plane (to the horizontal) and hence the magnitude of the normal force F_Normal is Mg cosθ. The frictional force F_fr is equal and opposite to the component (Mg sinθ) of the weight (of the block) parallel to the incline, as long as the block is at rest. If the block is just beginning to slide down, F_fr = μ_s Mg cosθ where μ_s is the coefficient of limiting friction (coefficient of static friction). If the block is in motion, F_fr = μ_k Mg cosθ where μ_k is the coefficient of kinetic friction. You need not show these details in the free body diagram. But you need to know these things in solving problems].

(3) A block of mass M sliding down an incline under gravity (No other external force applied):

The free body diagram will be the same as that shown in (2) above. But in this case the three forces are not balanced and so you have to show a smaller magnitude for the frictional force F_fr so that there is a net force acting down the plane to move the block

If you are asked take into account the viscous drag force F_D due to air, you have to show it along the direction of F_fr.

(4) A toy boat of mass m₁ carrying pebbles of mass m₂ floats on the surface of water in a jar:

In this case the object is the toy boat carrying glass beads. The total mass of the object is m₁ + m₂. Two forces act on the object. They are:

(i) The weight (m₁ + m₂)g of the object acting vertically downwards

(ii) The force of buoyancy F_Buoy acting vertically upwards.

You may represent the object by a small rectangle and indicate the weight and the force of buoyancy respectively by downward and upward arrows of equal length as indicated in FBD (i). Since the floating object is in equilibrium, the two vectors have to be equal and opposite and their lines of action must coincide.

[The weight acts through the centre of gravity of the object where as the force of buoyancy acts through the centre of buoyancy (centre of gravity of the displaced liquid). But you may ignore such details].

If you represent the object by a dot, things become quite simple and you get the free body diagram as shown in FBD (ii).

(5) An iron block of mass M suspended (from a rigid support) using a string and partially (or fully) submerged in a liquid.

In this case three forces act on the object (iron block). They are:

(i) The real weight Mg of the block acting vertically downwards

(ii) The force of buoyancy F_Buoy acting vertically upwards

(iii) The tension T in the string, acting vertically upwards

The free body diagram of the iron block can be drawn either as in FBD (i) or as in FBD (ii).

Since the iron block is in equilibrium, you have to draw the magnitudes of the forces such that magnitude of Mg = Magnitude of (F_Buoy + T).

[Free body diagrams you draw in your answer sheets are rough sketches and hence no examiner will expect you to show the exact magnitudes of the forces].

(6) A skier of mass M skiing down a frictionless incline with a drag force F_d due to air opposing her motion

Three forces act on the object (skier). They are:

(i) The weight Mg of the skier acting vertically downwards

(ii) The drag force F_D acting parallel to and up the incline

(ii) The normal force F_Normal exerted by the incline.

Usually you will be asked to represent the skier by a dot. You will draw the free body diagram of the skier as shown.

[If there is friction along the incline, you will have to show the frictional force also in same direction as that of the drag force].

(7) A cricket ball of mass m moving through air after being hit by a batsman (with negligible air resistance)

Since air resistance is negligible, the only force acting on the ball is its weight mg acting vertically downwards. The free body diagram is therefore very simple as shown in the adjoining figure.

[The direction of motion of the ball is immaterial in this simple case of negligible air resistance and you get the same free body diagram at all positions of the ball along its trajectory in air. But, if air resistance (drag force due to air) F_D is significant and is to be considered, you will have to show it directed opposite to the direction of motion of the ball. This direction will change continuously with time if the ball is following a parabolic path in air. If you are asked to draw the free body diagram at the instant the ball is at the highest point of its parabolic trajectory, the drag force will be horizontal and opposite to the direction of motion of the ball. But if the ball is projected vertically up, it will be momentarily at rest at the highest point and hence the viscous drag force will be zero there].

(8) Blocks A and B of masses M₁ and M₂ (M₁ > M₂) tied to the ends of a string (of negligible mass) passing over a frictionless fixed pulley, with block A resting on the ground and block B hanging in air

You are required to draw the free body diagrams of block A as well as block B.

The arrangement of the blocks and pulley is shown in the figure. The required free body diagrams are also shown.

Three forces act on block A. They are:

(i) The weight m₁g of the block acting vertically downwards

(ii) The normal force F_Normal exerted by the ground, acting vertically upwards

(iii) The tension T in the string, acting vertically upwards

The important thing to be noted in the case of block A is that the normal force F_Normal exerted on it by the ground is not of magnitude M₁g but is less than M₁g. The value is (M₁– M₂) g and the tension T in the string is M₂ g

Two forces act on block B. They are:

(i) The weight M₁g of the block acting vertically downwards

(ii) The tension T in the string, acting vertically upwards

[Since the same segment of string connects the two masses, the tension in the string has the same value].

The above situation can be suitably modified in your question paper. For instance, the block A can be a man of mass 70 kg standing on the ground and holding the end of the string and the block B can be a boy of mass 30 kg suspended from the other end of the string!

[If both blocks A and B were in air, the system of masses will move with an acceleration since there is a net driving force (M₁–M₂)g acting on the system. The free body diagram of block A will be similar to that of block B and the normal force F_Normal will be absent. The magnitude of the tension T in the string will be greater than M₁g, but less than M₁g].

(9) A tethered hydrogen balloon of mass M kept stationary in air. The mass of the rope is negligible.

You are required to draw the free body diagram of the balloon.

Three forces act on the balloon. They are:

(i) The weight Mg of the balloon, acting vertically downwards.

(ii) The force of buoyancy F_Buoy acting vertically upwards.

(iii) The tension T in the string, acting vertically downwards.

Since the mean density of the hydrogen balloon is less than that of air, the force of buoyancy will be greater than the weight of the balloon. The weight Mg of the balloon and the tension T in the rope together prevent the balloon from rising up.

The free body diagram of the balloon can be drawn as shown in FBD (i). If you were asked to represent the balloon by a dot, you would draw the free body diagram as shown in FBD (ii).