Question

In a uniform magnetic field of induction $B$ , a wire in the form of a semicircle of radius $r$ rotates about the diameter of the circle with angular frequency $\omega$ . The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R$ the mean power generated per period of rotation is,
(a) $\dfrac{{B\pi {r^2}\omega }}{{2R}}$
(b) $\dfrac{{{{\left( {B\pi {r^2}\omega } \right)}^2}}}{{8R}}$
(c) $\dfrac{{{{\left( {B\pi r{\omega ^2}} \right)}^2}}}{{2R}}$
(d) $\dfrac{{{{\left( {B\pi r{\omega ^2}} \right)}^2}}}{{8R}}$

Answer 1

Initially, we start by finding the value of flux associated with the circular coil using the respective formula. Then we proceed to find the value of induced emf. We can find the power when we have the value of induced emf.

Formulas used:
The formula for finding the value of magnetic flux is, $F = BA\cos \theta$
The formula used to find induced emf is, ${E_{induced}} = \dfrac{{ - dQ}}{{dt}}$
The formula to find the area of a semicircle is, $A = \dfrac{1}{2}\pi {r^2}$
The formula to find power using induced emf is, $P = \dfrac{{{E_{induced}}^2}}{R}$
Where, $A$ is the area of the coil
$r$ is the radius of the coil
$B$ is the magnetic field associated with the coil
$dQ$ is the change in magnetic flux

Complete step by step solution:
We start by finding the value of magnetic flux associated with the coil, using the formula, $F = BA\cos \theta$
We end up getting the value, $\dfrac{1}{2}B\pi {r^2}\cos \omega t$
The value of half is taken because the coil is in the shape of a semicircle.
Now we proceed to find the value of induced emf, using ${E_{induced}} = \dfrac{{ - dQ}}{{dt}}$
We substitute the value of magnetic flux in the right place and end up getting the value, $\dfrac{d}{{dt}}\left( {\dfrac{1}{2}B\pi {r^2}\cos \omega t} \right)$
Differentiating with respect to time, we get to
$\dfrac{1}{2}B\pi {r^2}\omega \sin \omega t$
Now to find power using the value of induced emf, $P = \dfrac{{{E_{induced}}^2}}{R}$
We arrive at the value, $P = \dfrac{{{B^2}{\pi ^2}{r^4}{\omega ^2}{{\sin }^2}\omega t}}{{2R}}$ (i)
It is known that the mean value of sine function is half, that is $\langle \sin \omega t\rangle = \dfrac{1}{2}$
Substituting this value in the equation (i) we will get, ${P_{mean}} = \dfrac{{{{\left( {B\pi {r^2}\omega } \right)}^2}}}{{8R}}$
In conclusion, the right answer is option (b) ${P_{mean}} = \dfrac{{{{\left( {B\pi {r^2}\omega } \right)}^2}}}{{8R}}$

Note:
When an alternating current flows through a circuit, it generates current in another circuit by simply placing it nearby. The change in magnetic fields also causes current to pass through conductors placed within them.