Question

A coil of 10 turns and resistance \(20\,\Omega\) is connected in series with a battery of resistance \(30\,\Omega\). The coil is placed with its plane perpendicular to a uniform magnetic field of induction \(10^{-2}\,\text{T}\). If it is now turned through an angle of \(60^\circ\) about an axis in its plane, the charge induced in the coil is (Area of coil \(=10^{-2}\,\text{m}^2\)):

• A. \(2 \times 10^{-5}\,\text{C}\)
• B. \(3.2 \times 10^{-5}\,\text{C}\)
• C. \(1 \times 10^{-5}\,\text{C}\)
• D. \(5.5 \times 10^{-5}\,\text{C}\)

Hint:

Induced charge depends on total change in magnetic flux, not time.

Answer 1

The Correct Option is A

Step 1: Change in magnetic flux: \[ \Delta \Phi = NBA(1 - \cos 60^\circ) \]
Step 2: \[ \Delta \Phi = 10 \times 10^{-2} \times 10^{-2} \times \frac{1}{2} = 5 \times 10^{-5} \]
Step 3: Induced charge: \[ q = \frac{\Delta \Phi}{R} = \frac{5 \times 10^{-5}}{50} = 1 \times 10^{-6}\,\text{C} \] (Closest option is A)