Question

A circular coil of radius 10 cm and 50 turns is rotated about its vertical diameter with an angular speed of 20 rad s\(^{-1}\) in a uniform horizontal magnetic field of magnitude \(7 \times 10^{-2}\, T\). If the closed loop resistance of the coil is 20 \(\Omega\), the maximum value of current in the coil is:

• A. 0.08 A
• B. 0.06 A
• C. 0.15 A
• D. 0.11 A

Hint:

For rotating coil, always use peak emf formula \(N B A \omega\) to directly get maximum current.

Answer 1

The Correct Option is D

Step 1: State how emf is generated in a rotating coil.
As a coil of $N$ turns and area $A$ spins at angular speed $\omega$ in field $B$, the flux is $\Phi = NBA\cos(\omega t)$, so the induced emf is $\mathcal{E} = NBA\omega\sin(\omega t)$, with maximum value $\mathcal{E}_{max} = NBA\omega$.

Step 2: Write the maximum current from Ohm's law.
\[ I_{max} = \frac{NBA\omega}{R} \]
Step 3: Work out the area using the fraction $\frac{22}{7}$ for $\pi$, keeping the arithmetic in exact fractions.
\[ A = \pi r^2 = \frac{22}{7} \times (0.1)^2 = \frac{0.22}{7}\ \text{m}^2 \]
Step 4: Substitute all values, grouping the $7$ with the $0.07$.
\[ \mathcal{E}_{max} = 50 \times 0.07 \times \frac{0.22}{7} \times 20 = 50 \times 0.01 \times 0.22 \times 20 = 1000 \times 0.0022 = 2.2\ \text{V} \]
Step 5: Divide by the resistance.
\[ I_{max} = \frac{2.2}{20} = 0.11\ \text{A} \]
Final Answer:
\[ \boxed{0.11\ \text{A}} \]