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 \(90\degree\) around any of its coplanar axis in \(0.1 s\), the emf induced in the coil will be:

• A. \(2V \)
• B. \(0.2V \)
• C. \(2×10^{-3}V \)
• D. \(0.02V \)

Answer 1

The Correct Option is D

To determine the induced electromotive force (emf) in the coil, we can use Faraday's Law of Electromagnetic Induction. According to Faraday's law, the induced emf (\(\varepsilon\)) in a coil is given by the rate of change of magnetic flux through the coil.

The formula for induced emf is:

\(\varepsilon = -\dfrac{\Delta \Phi}{\Delta t} \)

where:

• \(\Delta \Phi\) is the change in magnetic flux, and
• \(\Delta t\) is the time duration of the change.

The magnetic flux (\(\Phi\)) through the coil is given by:

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

where:

• \(B\) is the magnetic field,
• \(A\) is the area of the coil, and
• \(\theta\) 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 \(\theta = 0^\degree\) and \(\cos \theta = \cos 0^\degree = 1\).

After rotation by \(90^\degree\), the coil is parallel to the magnetic field, so \(\theta = 90^\degree\) and \(\cos \theta = \cos 90^\degree = 0\).

Thus, the initial magnetic flux is:

\(\Phi_i = B \cdot A \cdot 1 = 5 \times 10^{-5} \cdot 0.05 = 2.5 \times 10^{-6} \ Wb\)

The final magnetic flux is:

\(\Phi_f = B \cdot A \cdot 0 = 0 \ Wb\)

The change in magnetic flux is:

\(\Delta \Phi = \Phi_f - \Phi_i = 0 - 2.5 \times 10^{-6} = -2.5 \times 10^{-6} \ Wb\)

The induced emf is then calculated using the change in flux and the time duration of \(\Delta t = 0.1 \ s\):

\(\varepsilon = -\dfrac{\Delta \Phi}{\Delta t} = -\dfrac{-2.5 \times 10^{-6}}{0.1} = 2.5 \times 10^{-5} \ V\)

However, since there are 800 turns in the coil, the total induced emf is:

\(\varepsilon_{total} = N \cdot \varepsilon = 800 \times 2.5 \times 10^{-5} = 0.02 \ V\)

Therefore, the induced emf in the coil is \(0.02V\), which matches the given correct answer.