Question

A plane circular coil is rotated about its vertical diameter with a constant angular speed \( \omega \) in a uniform horizontal magnetic field. Initially the plane of the coil is parallel to the magnetic field. Draw plots showing the variation of the following physical quantities as a function of \( \omega t \), where \( t \) represents time elapsed: Magnetic flux \( \phi \) linked with the coil, and emf induced in the coil.

Hint:

In rotating coil problems:

Flux → sine or cosine depending on initial angle
emf is derivative of flux → phase difference \( 90^\circ \)
If flux starts from zero, emf starts from maximum.

Answer 1

Concept: Magnetic flux through a rotating coil: \[ \phi = BA \cos \theta \] Where:

\( \theta = \omega t \)
Coil rotates with angular speed \( \omega \)
Induced emf: \[ e = -\frac{d\phi}{dt} \]
Step 1: Initial condition. Given: Plane of coil initially parallel to magnetic field. So, angle between area vector and field = \( 90^\circ \). Hence: \[ \phi = 0 \text{ at } t = 0 \]
Step 2: Expression for magnetic flux. As the coil rotates: \[ \theta = \omega t + \frac{\pi}{2} \] So: \[ \phi = BA \cos\left(\omega t + \frac{\pi}{2}\right) = BA \sin(\omega t) \] Graph: Magnetic flux varies sinusoidally with time, starting from zero. So, \( \phi \) vs \( \omega t \) is a sine curve starting from origin.
Step 3: Induced emf. \[ e = -\frac{d\phi}{dt} = -BA\omega \cos(\omega t) \] Graph:

Cosine curve
Maximum at \( t = 0 \)
Phase difference of \( 90^\circ \) with flux

Step 4: Final Graph Description.

[(a)] \( \phi \) vs \( \omega t \): sine wave starting from zero.
[(b)] \( e \) vs \( \omega t \): cosine wave starting from maximum value.

Answer 2

Instead of substituting a phase-shifted angle directly into the flux formula, let's set up the general form of the flux and emf functions for a coil rotating in a uniform field, and then fix the unknown phase constant using the initial condition given in the question.

1. General form. For a plane coil of area \( A \) and \( N \) turns rotating with constant angular speed \( \omega \) in a uniform field \( B \), the magnetic flux linked with the coil at any instant always has the form \[ \phi(t) = BA\cos(\omega t + \delta), \] where \( \delta \) is a phase constant fixed by how the coil is oriented at \( t = 0 \).
2. Applying the initial condition. The question states that at \( t = 0 \) the plane of the coil is parallel to the magnetic field. When the coil's plane is parallel to \( \mathbf{B} \), its area vector, which is perpendicular to the plane, is itself perpendicular to \( \mathbf{B} \), so no field lines pass through the coil and the flux is zero at that instant: \[ \phi(0) = BA\cos\delta = 0 \implies \delta = \frac{\pi}{2}. \]
3. Flux as a function of time. Substituting \( \delta = \pi/2 \) and choosing the direction of the area-vector normal so the flux increases from zero as the coil begins to rotate into the field gives \[ \phi(t) = BA\sin(\omega t), \] a sine curve that starts at zero, rises to a maximum of \( BA \) at \( \omega t = \pi/2 \), and completes one full cycle as \( \omega t \) runs from \( 0 \) to \( 2\pi \).
4. Induced emf from Faraday's law. The emf is the negative rate of change of flux: \[ e(t) = -N\frac{d\phi}{dt} = -NBA\omega\cos(\omega t). \] This is a cosine curve, \( 90^\circ \) out of phase with the flux: it starts at its maximum magnitude at \( \omega t = 0 \), passes through zero exactly where the flux curve is at its own peak, and repeats periodically.

So the plot of \( \phi \) against \( \omega t \) is a sine wave starting from the origin, and the plot of \( e \) against \( \omega t \) is a cosine-shaped wave, shifted by a quarter cycle relative to the flux, reflecting that the emf is largest exactly when the flux is changing fastest, as the coil plane crosses the field direction, and zero when the flux is momentarily at its peak or trough.