Question

A 800 turn coil of effective area 0.05 $m^2$ is kept perpendicular to a magnetic field $5\times 10^{-5}$ T. When the plane of the coil is rotated by $90^\circ$ around any of its coplanar axis in 0.1 s, the emf induced in the coil will be

• A. $2\times 10^3$V
• B. 0.02 V
• C. 2V
• D. 0.2 V

Answer 1

The Correct Option is B

To determine the electromotive force (emf) induced in the coil, we will use Faraday's Law of Electromagnetic Induction, which states that the induced emf in a coil is given by the rate of change of the magnetic flux through the coil. The formula for emf (\( \mathcal{E} \)) is:

\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}

Where:

• N is the number of turns in the coil.
• \Delta \Phi is the change in magnetic flux.
• \Delta t is the time taken for the change.

The magnetic flux (\( \Phi \)) through a single turn of the coil is given by:

\Phi = B \cdot A \cdot \cos(\theta)

Where:

• B = 5 \times 10^{-5} \, \text{T} is the magnetic field strength.
• A = 0.05 \, \text{m}^2 is the effective area of the coil.
• \theta is the angle between the magnetic field and the normal to the plane of the coil.

Initially, the coil is perpendicular to the magnetic field, so \theta = 0^\circ, and thus:

\Phi_{\text{initial}} = 5 \times 10^{-5} \cdot 0.05 \cdot \cos(0^\circ) = 2.5 \times 10^{-6} \, \text{Wb}

After rotating the coil by \(90^\circ\), the angle becomes \(90^\circ\), and the new magnetic flux is:

\Phi_{\text{final}} = 5 \times 10^{-5} \cdot 0.05 \cdot \cos(90^\circ) = 0 \, \text{Wb}

The change in magnetic flux (\( \Delta \Phi \)) is:

\Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - 2.5 \times 10^{-6} = -2.5 \times 10^{-6} \, \text{Wb}

Substituting the values into Faraday's law, we have:

\mathcal{E} = -800 \cdot \frac{-2.5 \times 10^{-6}}{0.1} = 800 \cdot 2.5 \times 10^{-5} = 0.02 \, \text{V}

Therefore, the induced emf in the coil is 0.02 V.