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)].

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.

AP Physics C - Free Response Practice Question on Electromagnetic Induction

“Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages”

– Thomas A. Edison

Today I will give you a free response practice question on electromagnetic induction. You may try to answer this question within 15 minutes. Here is the question:

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.

This question carries 15 points. Try to answer it. I’ll be back soon with a model answer for your benefit.