Question

A rectangular coil of 200 turns and area $100\;cm^2$ is kept perpendicular to a uniform magnetic field of induction $0.25\;T$. If the field is reversed in direction in 0.01 second, the average induced emf in the coil is:
A. $10^6V$
B. $10^4V$
C. $10^2V$
D. Zero

Answer 1

Recall that the induced emf depends on the rate of change of the magnetic flux linked with the coil. Also, we see that the coil is perpendicular to the magnetic field which means that the magnetic field and the area vector are pointing in the same direction. Remember that the induced emf also depends on the number of turns in the coil. Plugging in the values into an established relation obtained from the preceding statements should lead you to the correct result.
Formula Used:
Induced emf $\epsilon = -N\dfrac{\Delta \phi}{\Delta t}$

Complete step by step answer:
Let us approach this problem in the perspective of Faraday’s Laws of Electromagnetic Induction.
Faraday’s first law suggests that whenever the magnetic flux linked with a coil changes, an emf is induced in the coil. The induced emf exists as long as the change in magnetic flux continues to exist. Recall that magnetic flux here is nothing but the magnetic field lines.
In the scenario presented to us, the change in the magnetic flux is brought about by reversing the direction of the magnetic field every 0.01 seconds.
Using Faraday’s first law we can say that this change in flux introduces an emf or a potential difference across the ends of the coil. This potential difference will induce a current that flows around the coil.
Now, Faraday’s second law of electromagnetic induction states that the magnitude of induced emf in a coil is equivalent to the rate of change of flux linked with the coil. This is quantised as:
$\epsilon = -N\dfrac{\Delta \phi}{\Delta t}$, where $\epsilon$ is the emf induced, $\phi$ is the magnetic flux, and $N$ is the number of turns in the coil. Note that the negative sign indicates that the direction of induced emf and the change in the direction of magnetic fields have opposite signs.
We know that the magnetic flux through the coil is given as $\phi = \vec{B}.\vec{A} = BAcos\theta$.
Since the coil is placed perpendicular to the magnetic field B, the area vector A is parallel to the magnetic field B $\Rightarrow \theta =0^{\circ} \Rightarrow cos\theta = cos 0^{\circ} = 1$
Therefore, $\phi = BA$
However, we are given that there is a reversal in field direction, which causes a change in flux through the coil:
$\Delta \phi = A\Delta B = A(B_f – B_i)$
Therefore, the induced emf $\epsilon = -N\dfrac{\Delta \phi}{\Delta t} = -N\dfrac{A(B_f-B_i)}{\Delta t}$
Substituting the values $N=200$, $A = 100\;cm^2 = 100 \times 10^{-4} = 10^{-2}\;m^2$, $B_f = -0.25T$, $B_i = 0.25T$ and $\Delta t = 0.01\;s$:
$\epsilon = -200\dfrac{10^{-2}(-0.25-0.25)}{0.01} = 10^{-2} \times 10^{4} = 10^{2}\;V$
Therefore, the correct choice is C. $10^2\;V$

Note:
Recall that the other ways to change the emf induced in a coil in addition to reversing the direction of the applied field are by:
1. Establishing relative motion between the field and the coil.
2. Changing the number of turns in the coil
3. By changing the strength of the magnetic field itself, which in turn changes the magnetic flux and consequently, the emf.