Question

What will happen when the frequency of rotation in an AC dynamo is doubled?

Answer 1

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Answer 2

Answer:

On doubling the frequency of rotation, the peak value of emf induced in the coil will also get doubled and so the peak value of induced current will also get doubled.

Explanation:

The magnetic flux linked with the coil of the AC dynamo is given by

$\phi = \vec B \cdot \vec A=BA\cos\theta$.

where,

• $\vec B$ = magnetic field passing through the coil of the dynamo.

• $\vec A$ = area vector of the coil which is directed along the normal to the plane of the coil.

If the coil is rotating with the angular frequency $\omega$, then the angle between $\vec B$ and $\vec A$ at an instant of time $t$ is given by

$\theta = \omega t$.

Also, $\omega = 2\pi ft$

$f$ = frequency of the rotation of the coil.

Therefore,

$\phi = BA\cos(\omega t)=BA\cos(2\pi ft)$.

which means, on doubling the frequency of rotation, the rate of change of the magnetic flux will also get doubled.

According to the Faraday's law of electromagnetic induction, the emf induced in the coil is given by,

$e=-\dfrac{\mathrm{d} \phi }{\mathrm{d} t}=-\dfrac{\mathrm{d} (BA\cos(2\pi ft))}{\mathrm{d} t}=BA(2\pi f) \sin(2\pi ft).$

Thus, on doubling the frequency the emf induced in the coil will also get doubled and so the induced current.