Question

A coil of area \( A \) and \( N \) turns is rotating with angular velocity \( \omega \) in a uniform magnetic field \( \mathbf{B} \) about an axis perpendicular to \( \mathbf{B} \). Magnetic flux \( \phi \) and induced emf \( \varepsilon \) across it, at an instant when \( \mathbf{B} \) is parallel to the plane of the coil, are:

• A. \( \phi = AB, \, \varepsilon = 0 \)
• B. \( \phi = 0, \, \varepsilon = 0 \)
• C. \( \phi = 0, \, \varepsilon = NAB\omega \)
• D. \( \phi = AB, \, \varepsilon = NAB\omega \)

Hint:

When dealing with rotating coils in a magnetic field, use Faraday's Law to calculate the induced emf and remember that the magnetic flux depends on the angle between the magnetic field and the normal to the coil. The induced emf depends on the rate of change of magnetic flux.

Answer 1

The Correct Option is D

The magnetic flux \( \phi \) is given by the product of the magnetic field \( \mathbf{B} \), the area \( A \) of the coil, and the cosine of the angle between the magnetic field and the normal to the plane of the coil: \[ \phi = NBA \cos(\theta) \] At the instant when \( \mathbf{B} \) is parallel to the plane of the coil, the angle \( \theta = 90^\circ \), so \( \cos(90^\circ) = 0 \). Thus, the magnetic flux \( \phi = AB \). The induced emf \( \varepsilon \) can be found using Faraday's Law of Induction, which states that the induced emf is equal to the rate of change of the magnetic flux: \[ \varepsilon = -\frac{d\phi}{dt} \] Since the coil is rotating with angular velocity \( \omega \), the rate of change of the flux is \( NAB\omega \), so the induced emf is \( \varepsilon = NAB\omega \). Thus, the correct answer is \( \phi = AB \) and \( \varepsilon = NAB\omega \).