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 B & C - Multiple Choice Practice Questions on Electromagnetic Induction

“Maturity is often more absurd than youth and very frequently is more unjust to youth.”

– Thomas A. Edison

Michael Faraday’s discovery of electromagnetic induction was a turning point in the history of mankind. When he made the first public announcement that the relative motion between a magnet and a coil of wire could cause the flow of a feeble electric current through the coil, he had to face this question: “But what is the use?” Faraday countered this with another question: “What is the use of a new born baby?”

The baby has grown rapidly to become a very healthy youth who will remain so for many more decades!

The phenomenon responsible for the generation of electric power for feeding the modern world still continues to be electromagnetic induction.

Questions on electromagnetic induction are generally interesting. Click on the label ‘electromagnetic induction’ below this post; you will find all posts on electromagnetic induction published so far on this site.

Today we will discuss a few more multiple choice practice questions in this section.

(1) Earth’s resultant magnetic field at California has magnitude B tesla and it makes an angle θ with the horizontal. Assuming that there are no other magnetic fields, what will be the voltage induced between the tips of the wings of an airplane of wing-span L flying horizontally with speed v?

(a) BLv

(b) BLv sin θ

(c) BLv cos θ

(d) BLv/sin θ

(e) BLv/cos θ

Since the airplane is flying horizontally it can ‘cut’ the vertical magnetic field lines to generate a motional emf V given by

V = B_verticalLv where B_vertical is the vertical component of earth’s magnetic field at the place.

With reference to the adjoining figure we have

B_vertical = B sin θ

Therefore, the voltage induced between the tips of the wings of the airplane is BLv sin θ.

(2) A plane square loop of thin copper wire has 100 turns. Each side of the loop is 10 cm long and the loop is oriented with its plane making an angle of 30º with a uniform magnetic field of flux density 0.4 tesla. If the loop is rotated in 0.5 second so as to orient its plane at right angles to the magnetic field, what will be the magnitude of the average emf induced in the loop?

(a) 0.1 volt

(b) 0.2 volt

(c) 0.4 volt

(d) 0.8 volt

(e) 2 volt

The induced emf V is given by

V = – dФ/dt where dФ is the change in the total magnetic flux linked with the coil and dt is the time taken for the flux change.

[The negative sign is the consequence of Lenz’s law by which the induced emf has to oppose the change of flux dФ].

Since we are required just to find the magnitude of the induced voltage, we may ignore the negative sign

Since the coil has N (=100) turns, the total flux linked with the coil is Nφ where φ is the flux per turn given by

φ = BAcos θ where B = 0.4 tesla and A = area of the square loop = (0.1)² m² = 0.01 m²

The angle θ is the angle between the magnetic field and the area vector.

[Remember that the area vector is directed perpendicular to the plane of the coil].

Since the plane of the coil makes an angle of 30º with a magnetic field, the area vector makes an angle of 60º with the magnetic field.

The initial magnetic flux linkage is NBAcos 60º = 100×0.4×0.01×(1/2) = 0.2 weber.

Since the area vector and the magnetic field are finally parallel (or anti-parallel), the final flux linkage is NBAcos 0º = 100×0.4×0.01 = 0.4

The change of flux dФ = 0.4 – 0.2 = 0.2

Therefore, induced emf = (Change of flux) /(Time) = 0.2/0.5 = 0.4 volt.

(3) Suppose that the resistance (R) of the loop in the above question is 10 Ω. What will be the induced current in the loop if the loop is kept stationary and the magnetic field is steadily reduced to zero in a time of 40 millisecond?

(a) 0.2 A

(b) 0.5 A

(c) 1 A

(d) 1.5 A

(e) 2 A

The initial magnetic flux linked with the loop (as shown above) is NBAcos 60º = 100×0.4×0.01×(1/2) = 0.2 weber.

When the magnetic field is reduced to zero, the magnetic flux is reduced to zero. Therefore the change of magnetic flux is 0.2 weber. The emf V induced in the loop is given by

V = (Change of flux) /(Time) = 0.2/(40×10^–3) volt = 5 volt.

The current induced in the loop is V/R = 5/10 A = 0.5 A [Option (b)]

The following question is meant specifically for AP Physics C aspirants:

(4) A straight conductor of length L and mass M can slide down along a pair of long, smooth, conducting vertical rails P and Q of negligible resistance (Fig.). A resistor of resistance R is connected between the ends of the rails as shown in the figure. A uniform magnetic field of flux density B acts perpendicularly into to the plane containing the rails and the sliding conductor. The terminal velocity of fall of the rod is

(a) MgR/LB

(b) mgL/B²R²

(c) B²L²/mgR

(d) mgB/L²R

(e) mgR/B²L²

When the rod slides down under gravity, the magnetic flux linked with the closed circuit comprising the rod, rails and the resistor R changes and a current is induced in the circuit. The induced emf is the motional emf BLv where v is the velocity of the rod. The induced current I in the circuit is BLv/R.

By Lenz’s law the induced current has to oppose the motion of the rod. It is the magnetic force ILB which brings in this opposition. When the velocity of the rod increases, the opposing magnetic force also increases. When the magnitudes of the gravitational force (weight Mg of the rod) and the opposing magnetic force become equal, the rod moves with a constant (terminal) velocity.

Therefore, we have

ILB = Mg

Substituting for I we have (BLv/R)LB = Mg

Or, B²L²v/R = Mg

This gives v = MgR/B²L²

Now, let me ask you a question:

If the direction of the magnetic field in the above question is reversed, will the rod still attain a terminal velocity? Think of it and arrive at the answer ‘YES’.

You will find a few more questions (with solution) in this section here.

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)].