Question

A circular coil of radius 8 cm, 400 turns and resistance \( 2\,\Omega \) is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through \( 180^{\circ} \) in 0.30 sec. Horizontal component of the earth's magnetic field at the place is \( 3\times10^{-5} \) T. The magnitude of current induced in the coil is approximately:

• A. \( 4\times10^{-2}~\text{A} \)
• B. \( 8\times10^{-4}~\text{A} \)
• C. \( 8\times10^{-2}~\text{A} \)
• D. \( 1.92\times10^{-3}~\text{A} \)

Hint:

Flipping an induction loop coil completely upside down (\( 180^\circ \)) inside a uniform magnetic field always doubles the total flux change (\( 2NBA \)), creating a highly predictable value structure.

Answer 1

The Correct Option is B

Step 1: Pick Faraday's law.
A changing magnetic flux through the coil makes an induced emf. The emf equals the number of turns times how fast the flux changes: \[ e = \frac{N A B\,|\cos\theta_1 - \cos\theta_2|}{\Delta t} \]
Step 2: Find the coil area.
With radius $r = 0.08$ m: \[ A = \pi r^{2} = \pi(0.08)^{2} \approx 2.01\times10^{-2}\ \text{m}^{2} \]
Step 3: Find the flux change factor.
The coil flips by $180^{\circ}$, so the angle goes from $0^{\circ}$ to $180^{\circ}$. Then $\cos 0^{\circ} - \cos 180^{\circ} = 1 - (-1) = 2$. So the flux change is $2NBA$.
Step 4: Compute the emf.
With $N = 400$, $B = 3\times10^{-5}$ T, $\Delta t = 0.30$ s: \[ |e| = \frac{2(400)(3\times10^{-5})(2.01\times10^{-2})}{0.30} \approx 1.6\times10^{-3}\ \text{V} \]
Step 5: Use Ohm's law for current.
With resistance $R = 2\ \Omega$: \[ I = \frac{|e|}{R} = \frac{1.6\times10^{-3}}{2} \]
Step 6: Get the final current.
\[ I \approx 8\times10^{-4}\ \text{A} \] \[ \boxed{I \approx 8\times10^{-4}\ \text{A}} \]