Question

A circular coil of area \(3 \times 10^{-2} \text{ m}^2\), 900 turns and a resistance of 1.8 Ω is placed with its plane perpendicular to a uniform magnetic field of \(3.5 \times 10^{-5}\) T. The current induced in the coil when it is rotated through 180° in half a second is

• A. 2.1 mA
• B. 1.8 mA
• C. 1.5 mA
• D. 2.7 mA

Hint:

When a coil is rotated by 180° from a position where its plane is perpendicular to the field, the change in flux is always \(2BA\). When it's rotated by 90°, the change is \(BA\). Memorizing these common cases can save time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem is based on Faraday's Law of Electromagnetic Induction. When the magnetic flux through a coil changes, an electromotive force (EMF) is induced, which in turn drives a current if the circuit is closed.
Step 2: Key Formula or Approach:
1. Magnetic Flux (\(\Phi\)): \(\Phi = BA \cos\theta\), where \(\theta\) is the angle between the magnetic field vector \(\vec{B}\) and the area vector \(\vec{A}\).
2. Induced EMF (\(\mathcal{E}\)): According to Faraday's Law, \(\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}\), where \(N\) is the number of turns and \(\Delta \Phi\) is the change in flux over time \(\Delta t\).
3. Induced Current (\(I\)): Using Ohm's Law, \(I = \frac{|\mathcal{E}|}{R}\), where \(R\) is the resistance of the coil.
Step 3: Detailed Explanation:
Given values:
- Area, \(A = 3 \times 10^{-2} \text{ m}^2\).
- Number of turns, \(N = 900\).
- Resistance, \(R = 1.8 \text{ Ω}\).
- Magnetic field, \(B = 3.5 \times 10^{-5} \text{ T}\).
- Time interval, \(\Delta t = 0.5 \text{ s}\).
Calculate the change in magnetic flux (\(\Delta \Phi\)):
- Initially, the plane is perpendicular to the field. This means the area vector \(\vec{A}\) is parallel to \(\vec{B}\). So, the initial angle is \(\theta_i = 0^\circ\).
- Initial Flux: \(\Phi_i = BA \cos(0^\circ) = BA\).
- The coil is rotated through 180°. So, the final angle is \(\theta_f = 180^\circ\).
- Final Flux: \(\Phi_f = BA \cos(180^\circ) = -BA\).
- Change in Flux: \(\Delta \Phi = \Phi_f - \Phi_i = -BA - BA = -2BA\).
Calculate the induced EMF (\(|\mathcal{E}|\)):
\[ |\mathcal{E}| = N \left| \frac{\Delta \Phi}{\Delta t} \right| = N \frac{2BA}{\Delta t} \] \[ |\mathcal{E}| = 900 \times \frac{2 \times (3.5 \times 10^{-5} \text{ T}) \times (3 \times 10^{-2} \text{ m}^2)}{0.5 \text{ s}} \] \[ |\mathcal{E}| = 900 \times \frac{21 \times 10^{-7}}{0.5} = 900 \times 42 \times 10^{-7} = 37800 \times 10^{-7} = 3.78 \times 10^{-3} \text{ V} \] Calculate the induced current (\(I\)):
\[ I = \frac{|\mathcal{E}|}{R} = \frac{3.78 \times 10^{-3} \text{ V}}{1.8 \text{ Ω}} = 2.1 \times 10^{-3} \text{ A} \] Converting to milliamperes (mA):
\[ I = 2.1 \text{ mA} \] Step 4: Final Answer:
The induced current is 2.1 mA. Therefore, option (A) is correct.