Question

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

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

Hint:

When a coil rotates in a magnetic field, the flux and induced emf depend on the orientation of the magnetic field and the angular velocity of rotation.

Answer 1

The Correct Option is D

The problem involves finding the magnetic flux \(\varphi\) and the induced electromotive force (emf) \(\varepsilon\) in a rotating coil within a magnetic field.

Here's a step-by-step analysis:

Magnetic Flux \(\varphi\):

The magnetic flux through a coil is given by \(\varphi = B \cdot A \cdot \cos(\theta)\), 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.

When the magnetic field \(\vec{B}\) is parallel to the plane of the coil, \(\theta = 90^\circ\), hence, the flux \(\varphi = B \cdot A \cdot \cos(90^\circ) = 0\).

Induced emf \(\varepsilon\):

The induced emf in a rotating coil is given by Faraday’s law of electromagnetic induction, \(\varepsilon = -\frac{d\varphi}{dt}\).

For a coil with area \(A\), \(N\) turns, and rotating with angular velocity \(\omega\) in a magnetic field \(B\), when the magnetic field is parallel to the plane of the coil, the rate of change of flux is maximal.

The flux \(\varphi = B \cdot A \cdot \cos(\theta) = B \cdot A \cdot \cos(\omega t)\).

Differentiating with respect to time \(t\), we get \(\varepsilon = -N \frac{d}{dt}(BA\cos(\omega t)) = NAB\omega\sin(\omega t)\).

At the instant when \(\theta = 90^\circ\), \(\sin(\omega t) = 1\). Thus, \(\varepsilon = NAB\omega\).

Thus, at the moment \(\vec{B}\) is parallel to the coil plane, we have:

Result:

\(\varphi = 0, \quad \varepsilon = NAB\omega\)