“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*_{vertical}*Lv* 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* = – d*Ф*/d*t* where d*Ф *is the change in the *total* magnetic flux linked with the coil and d*t* 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 d*Ф*].

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 = *BA*cos* θ* where *B* = 0.4 tesla and *A = *area of the square loop = (0.1)^{2} m^{2} = 0.01 m^{2}

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 *N**BA*cos* *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 N*BA*cos* *0º = 100×0.4×0.01 = 0.4

The change of flux d*Ф *= 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 *N**BA*cos* *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/B*^{2}*R*^{2}

(c) *B*^{2}*L*^{2}*/mgR*

(d) *mgB/L*^{2}*R*

(e) *mgR/B*^{2}*L*^{2}* *

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, *B*^{2}*L*^{2}*v/R = Mg*

This gives *v = MgR/B*^{2}*L*^{2}

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.

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