AP Physics Multiple Choice Practice Questions on Electromagnetic Induction

“There is no democracy in physics. We can't say that some second-rate guy has as much right to an opinion as Fermi.”

–Luis Walter Alvarez

Today we shall discuss a few questions involving electromagnetic induction. Questions in this section are generally interesting and we have discussed many typical questions on various occasions on this site.

(1) A single turn plane circular conducting loop of area A and resistance R is placed in a uniform magnetic field of flux density B which has a time rate of change. The plane of the loop is perpendicular to the magnetic field. If the emf induced in the loop is V, the time rate of change of the magnetic flux density is

(a) V/A

(b) V/RA

(c) RV/A

(d) A/V

(e) AR/V

The induced emf V is  given by

             V = ∆Φ/∆t  where ∆Φ is the change of magnetic flux occurring in a small time ∆t.  

Since Φ = BA we have

             V = ∆BA/∆t  

The time rate of change of the magnetic flux density is ∆B/∆t = V/A

(2) A long straight power line carries a current I which decreases with time at a uniform rate. A plane circular conducting loop is arranged below the power line as shown in the figure. Which one among the following statements is true?

(a) No current is induced in the circular loop.

(b) A uniformly decreasing current is induced in the loop.

(c) A uniformly increasing current is induced in the loop.

(d) A steady current is induced in the loop and it flows in the anticlockwise direction

(e) A steady current is induced in the loop and it flows in the clockwise direction

Since the current in the power line is changing, the magnetic flux linked with the circular loop is changing. Therefore there must be an induced current in the loop. The induced current in the loop must be steady since the rate of decrease of magnetic flux is steady (because of the uniform decrease of current in the power line).

The magnetic field lines produced by the current in the power line are directed normally into the plane of the loop. Since the current in the power line decreases with time, the induced current in the loop must supply magnetic flux lines in the same direction, in accordance with Lenz’s law (for opposing the reduction of the flux). Therefore the induced current in the circular loop must flow in the clockwise direction [Option (e)]. 

The following questions are meant for AP Physics C aspirants:

(3) Two horizontal conducting rails AB and CD of negligible resistance are connected by a conductor BC of resistance R. Another conducting rod PQ of length L and negligible resistance can slide without friction along the rails (Fig.). The plane ABCD is horizontal and a constant magnetic field B tesla acts perpendicular to the plane ABCD. A small constant horizontal force F is applied on the slider PQ perpendicular to its length so that it slides with a constant velocity ‘v’. What is the value of the velocity v?

(a) FR/BL

(b) FR/B²L²

(c) FR/B²L

(d) FR/BL²

(e) FR²/B²L²

On applying the force F, the rod PQ starts to move from rest with an acceleration. When the rod moves the magnetic flux linked with the circuit PBCQ changes and an emf is induced in the circuit. Obviously this is the motional emf BLv.

[Note that when a conductor of length L moves with velocity ‘v’ at right angles to a magnetic field of flux density B, the motional voltage generated between its ends is BLv].

Since we have a closed circuit PBCQ, the emf BLv drives a current ‘I’ through it. PQ is therefore a current carrying conductor moving at right angles to a magnetic field. A magnetic force ILB acts opposite to the direction of motion of the conductor (in accordance with Lenz’s law). The opposing magnetic force goes on increasing with the increase in velocity of the conductor until the magnitude of the magnetic force becomes equal to that of the applied force F. The conductor thereafter continues to move with the terminal velocity acquired by it. The velocity of the rod after the initial accelerated motion is now constant.

Equating the magnitudes of the applied force F and the magnetic force ILB we have

             F = ILB

But I = BLv/R

Therefore F = B²L²v/R

This gives v = FR/B²L²

(4) An inductance L and a resistance R are connected in series with a battery and switch S as shown in the figure. The switch is closed at time t = 0. Which one among the following graphs gives the variation of the voltage V_L across the inductance as a function of time t?

There will be a voltage drop across the inductance only if the current in it changes. When the switch is closed the current in the series LR circuit will rise rapidly initially and will finally settle at the final maximum value.

[The final maximum current I₀ in the LR circuit is V₀/R where V₀ is the emf of the battery. The current I in the LR circuit during the growth at any instant t is given by

             I = I₀[1 – e^–Rt/L] where e is the base of natural logarithms]

Since the rate of variation of current is maximum initially, the voltage induced in the inductance is maximum initially. The rate of variation of current is non-linear and finally becomes zero. Therefore, the emf induced in the inductance finally becomes zero and the variation of the voltage V_L across the inductance as a function of time t is correctly represented by graph (b).

[Note that graph (d) is incorrect since the variation of the voltage is linear].

You can access all posts on electromagnetic induction on this site by clicking on the label ‘electromagnetic induction’ below this post.

 

AP Physics C – Additional Practice Questions (MCQ) on Electromagnetic Induction

"There's a way to do it better - find it"
– Thomas A. Edison

In my last post I had given you a few multiple choice practice questions (with solution) involving electromagnetic induction. Today we will discuss a few more questions in this section. These are meant for AP Physics C aspirants even though AP Physics B aspirants also will find them useful:

(1) The adjoining figure shows a 1000 turn coil of fine insulated copper wire connected to the y-inputs A and B of a cathode ray oscilloscope set for displaying the voltage wave form induced in the coil. A bar magnet, with its axis vertical and coinciding with the axis of the coil, is initially at rest, with its centre O at a height 10 cm from the centre of the coil. The bar magnet is allowed to fall freely under gravity and the induced voltage as a function of time is displayed on the screen of the oscilloscope. Which one among the following graphs best represents the induced voltage?

The magnetic flux linked with the coil increases up to the instant when the centres of the magnet and the coil coincide. Thereafter the magnetic flux decreases. Therefore, the direction of the induced emf gets reversed. The reversal of the induced emf occurs at the instant when the centres of the magnet and the coil coincide since the rate of change of magnetic flux at that instant is zero even though the flux linkage is maximum. Further, the latter half of the induced voltage has a peak of greater magnitude since the speed of fall of the magnet is greater so that the rate of change of magnetic flux is greater. The correct option is (a).

(2) The voltage variation across a resistance R in a series LR circuit is displayed as a function of time using a cathode ray oscilloscope. The swith S is closed at time t = 0 and then opened at time t = t₁. Which one among the following graphs best represents the voltage variation across R during the time interval 0 to t₁?

When the switch S is closed,the time constant of the circuit is L/R and is significant. The current I in the circuit therefore does not rise abruptly to the final maximum value I₀ (let us say). . The rise of current is exponential and hence the voltage across the resistance R rises exponentially with time.

[The exponential growh of the current I is given by I = I₀(1 – e ^–Rt/L) where e is the base of natural logarithms].

When the switch S is opened at the instant t₁ the open circuit at the switch makes the resistance of the circuit infinite and hence the time constant becomes zero. The current drops abruptly to zero. The voltage drop across R also drops abruptly to zero. The correct option is (e).

[The graph (d) may distract you. The growth of the voltage shown in it is not exponential. In an exponential growth, the initial rate of growth will be the largest].

(3) The switch S in the circuit shown is closed and sufficient time is allowed so that the current through the inductance L and the resistance R becomes the final steady value. The inductance L is about 20 henry and the resistance R is about 10 ohm. The switch S is opened at time t = 0. Then the voltmeters V₁ and V₂ will indicate the same reading at

(a) time t = 0

(b) time t = L/R

(c) rime t = 2 L/R

(d) time t = L/2R

(e) all times

This question is very simple but it may confuse you.

Since the LR circuit has a non-zero time constant (L/R), the voltmeter readings will not become zero abruptly but will drop to zero exponentially with time. Since the inductance and resistance are connected in parallel, the voltmeters V₁ and V₂ will indicate the same reading at all times [Option (e)].

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 Electromagnetic Induction

A free response practice question on electromagnetic induction was posted on 10^th October 2010. As promised, I give below a model answer (along with the question) for your benefit.

[You can access earlier posts in this section either by clicking on the label 'electromagnetic induction' below this post or by trying a search for 'electromagnetic induction' using the search box provided on this page].

Two infinitely long straight parallel wires W₁ and W₂, separated by a distance ‘a’ in free space, carry equal currents I flowing in opposite directions as shown in the adjoining figure. A square loop PQRS of side ‘a’, made of nichrome wire of resistance ρ Ω per metre is arranged with its plane lying in the plane of the wires W₁ and W₂ so that the sides PQ and RS of the loop are parallel to the wires W₁ and W₂. The side PQ of the loop is at a distance ‘a’ from the wire W₂. Now, answer the following questions in terms of the given quantities and fundamental constants:

(a) Determine the magnetic flux density at a point midway between the wires W₁ and W₂.

(b) Determine the magnetic flux density at the mid point of the square loop PQRS.

(c) Calculate the magnetic flux through the loop PQRS.

(d) What is the average emf induced in the loop when the current through the wires is switched off in a time of 50 ms?

(e) When the current through the wires is switched off, it is found that at a certain instant t, the current decays at the rate of 40 As^–1. Calculate the current induced in the loop PQRS at the instant t.

Indicate the direction of the current in the loop and justify your answer.

The magnitude of the magnetic flux density B at a point distant a from an infinitely long straight conductor carrying current I is given by

B = μ₀I/2πa where μ₀ is the magnetic permeability of free space

The magnetic flux density at a point midway between the wires W₁ and W₂ is the resultant of the magnetic flux densities produced by these wires. Each wire produces magnetic flux density of magnitude μ₀I/2πa.

These fields are directed perpendicular to the plane containing the wires and outwards (towards the reader) and hence they add up to produce a resultant flux density of magnitude μ₀I/2πa + μ₀I/2πa= μ₀I/πa.

[Since the magnetic flux density is a vector, you should not forget to mention its direction].

(b) At the mid point of the square loop PQRS the magnetic fields due to the wires W₁ and W₂ are directed perpendicular to the plane containing the wires. But the field due to the wire W₂ is directed into the plane of the figure (away from the reader) where as the field due to the wire W₁ is directed outwards (towards the reader). The resultant field is directed into the plane of the figure (away from the reader) since the wire W₂ produces stronger field.

The magnitude of the resultant magnetic flux density at the mid point of the square loop PQRS is [(μ₀I)/(2π×3a/2)] – [(μ₀I)/(2π×5a/2)]

This is equal to (μ₀I/πa)[(1/3) – (1/5)] = 2μ₀I/15πa

(c) To find the magnetic flux through the loop PQRS, consider a strip of very small width dx at distance x from the wire W₂ as shown in the figure. The resultant magnetic flux density B at the strip is given by

B = (μ₀I/ 2π) [1/x – 1/(a+x)]

Magnetic flux linked with the strip = B dA = Badx

[dA is the area of the strip of length a and width dx]

Magnetic flux Ф linked with the entire loop PQRS is given by

Ф = _a∫^2a Badx = (μ₀Ia/2π) _a∫^2a [1/x – 1/(a+x)]dx

[The limits of integration are x = a and x = 2a].

Therefore, Ф = (μ₀Ia/2π) [ln x – ln (a+x)] between limits x = a and x = 2a.

Or, Ф = = (μ₀Ia/2π) ln [x/(a+x)] between limits x = a and x = 2a.

= (μ₀Ia/2π) [ln(2/3) – ln(1/2)]

= (μ₀Ia /2π) ln (4/3) ...............................(i)

[The unit of magnetic flux density is tesla (or, weber per mrtre²) and the unit of magnetic flux is weber].

(d) The average emf V_average induced in the loop is given by

V_average = Rate of change of magnetic flux

= Change of magnetic flux/ Time

= [(μ₀Ia /2π) ln(4/3) – 0]/(50×10^–3) since the current through the wires is switched off in a time of 50 ms

Therefore, V_average = = (10μ₀Ia /π) ln (4/3)

(e) The emf V induced in the loop PQRS at the instant t is given by

V = – dФ/dt, the negative sign appearing because of Lenz’s law.

Ignoring the negative sign, we have from equation (i)

V = dФ/dt = [(μ₀a /2π) ln (4/3)](dI/dt)

Here dI/dt = 40 As^–1, as given in the question.

Substituting for dI/dt, we have V = (20μ₀a /π) ln (4/3)

The induced current in the loop = V/R where R is the resistance of the loop, which is equal to 4aρ ohm.

Therefore, induced current = (20μ₀a /4πaρ) ln (4/3) = (5μ₀ /πρ) ln (4/3) ampere.

The direction of the induced current in the loop is clockwise, as indicated in the figure.

The resultant magnetic field is directed normally into the plane of the loop and is decreasing when the current is switched off. The induced current should oppose this change and should therefore produce a magnetic field acting in the same direction (normally into the plane of the loop). This is made possible by the clockwise flow of the induced current.