Question

A coil has an area $0.06\text{ m}^2$ and it has $600$ turns. After placing the coil in a magnetic field of strength $5 \times 10^{-5}\text{ Wb/m}^2$, it is rotated through $90^\circ$ in $0.2$ second. The magnitude of average e.m.f induced in the coil is [$\cos 0^\circ = \sin 90^\circ = 1$ and $\sin 0^\circ = \cos 90^\circ = 0$]

• A. $3\text{ mV}$
• B. $9 \times 10^{-3}\text{ V}$
• C. $12 \times 10^{-3}\text{ V}$
• D. $3\text{ V}$

Hint:

To simplify calculations with decimals, group the numbers into whole factors first: $\frac{600 \times 0.06}{0.2} = \frac{36}{0.2} = 180$. Then simply multiply this combined factor by the remaining magnetic field term: $180 \times 5 \times 10^{-5} = 900 \times 10^{-5} = 9 \times 10^{-3}\text{ V}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A multi-turn conducting coil ($N = 600$, $A = 0.06\text{ m}^2$) is initially positioned perpendicular to a uniform magnetic field ($B = 5 \times 10^{-5}\text{ Wb/m}^2$).
The coil is then rotated by $90^\circ$ over a time interval of $\Delta t = 0.2\text{ s}$. We need to compute the magnitude of the average electromotive force (e.m.f.) induced across the coil terminals.

Step 2: Key Formula or Approach:
According to Faraday's Law of Electromagnetic Induction, the average induced e.m.f. ($e$) corresponds to the time rate of change of the linked magnetic flux:
$$|e| = N \frac{|\Delta \phi|}{\Delta t} = N \frac{|\phi_2 - \phi_1|}{\Delta t}$$ The magnetic flux linked through a single loop depends on its orientation: $\phi = BA \cos\theta$.
Initially, the coil faces the flux directly ($\theta_1 = 0^\circ$), and it rotates to a parallel orientation ($\theta_2 = 90^\circ$).

Step 3: Detailed Explanation:
Let's evaluate the initial ($\phi_1$) and final ($\phi_2$) magnetic flux states:
$$\phi_1 = BA \cos(0^\circ) = BA(1) = BA$$ $$\phi_2 = BA \cos(90^\circ) = BA(0) = 0$$ The net change in magnetic flux per turn is:
$$|\Delta \phi| = |\phi_2 - \phi_1| = |0 - BA| = BA$$ Now, substitute this flux difference into Faraday's equation along with the system variables:
$$|e| = \frac{N \cdot B \cdot A}{\Delta t}$$ Substitute the given values ($N = 600$, $B = 5 \times 10^{-5}$, $A = 0.06$, $\Delta t = 0.2$):
$$|e| = \frac{600 \times (5 \times 10^{-5}) \times 0.06}{0.2}$$ Simplify the numerical multiplication in the numerator:
$$600 \times 0.06 = 36$$ $$36 \times 5 \times 10^{-5} = 180 \times 10^{-5} = 1.8 \times 10^{-3}$$ Now divide by the time denominator:
$$|e| = \frac{1.8 \times 10^{-3}}{0.2} = 9 \times 10^{-3}\text{ V}$$

Step 4: Final Answer:
The magnitude of the average induced e.m.f. is $9 \times 10^{-3}\text{ V}$, which corresponds to option (B).