Question

A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.010 T exists in a direction perpendicular to the axis of rotation. Find (a) the maximum emf induced, (b) the average emf induced in the coil over a long period and (c) the average of the squares of emf induced over a long period.

Answer 1

Given,
Radius of the circular coil, R = 5.0 cm
Angular speed of circular coil, $\omega$ = 80 revolutions/minute
Magnetic field acting perpendicular to the axis of rotation, B = 0.010 T
The emf induced in the coil $\left(e\right)$ is given by,
$$\begin{gathered} e=\frac{d\varphi }{dt} \\ \Rightarrow e=\frac{dB.A\cos \theta }{dt} \\ \Rightarrow e=-BA\sin \theta \frac{d\theta }{dt} \end{gathered}$$
$\Rightarrow$e = −BAωsinθ

($\frac{d\theta }{dt}=\omega$ = the rate of change of angle between the arc vector and B)

(a) For maximum emf, sinθ = 1
$\therefore$ e = BAω
$\Rightarrow$e = 0.010 × 25 × 10⁻⁴ × 80 × $\frac{2\mathrm{\pi }\times \mathrm{\pi }}{60}$
$\Rightarrow$e = 0.66 × 10⁻³ = 6.66 × 10⁻⁴ V

(b) The direction of the induced emf changes every instant. Thus, the average emf becomes zero.

(c) The emf induced in the coil is e = −BAωsinθ = −BAωsin ωt

The average of the squares of emf induced is given by
$$\begin{gathered} {e_{\mathrm{av}}}^2=\frac{{\int }_0^T{B}^2{A}^2{\omega }^2{\sin }^2\omega t\mathrm{d}t}{{\int }_0^T\mathrm{d}t} \\ \Rightarrow {e_{\mathrm{av}}}^2=\frac{B^2{A}^2{\omega }^2{\int }_0^T{\sin }^2\omega t\mathrm{d}t}{{\int }_0^T\mathrm{d}t} \\ \Rightarrow {e_{\mathrm{av}}}^2=\frac{B^2{A}^2{\omega }^2{\int }_0^T\left(1-\cos 2\omega t\right)\mathrm{d}t}{2T} \\ \Rightarrow {e_{\mathrm{av}}}^2=\frac{B^2{A}^2{\omega }^2}{2T}{\left[t-\frac{\sin 2\omega t}{2\omega }\right]}_0^T \\ \Rightarrow {e_{\mathrm{av}}}^2=\frac{B^2{A}^2{\omega }^2}{2T}\left[T-\frac{\sin 4\pi -\sin 0}{2\omega }\right]=\frac{B^2{A}^2{\omega }^2}{2} \\ \Rightarrow {e_{\mathrm{av}}}^2=\frac{(6.66\times {10}^{-4}{)}^2}{2}=22.1778\times {10}^{-8}{\mathrm{V}}^2\left[\because BA\omega =6.66\times {10}^{-4}\mathrm{V}\right] \\ \Rightarrow {e_{\mathrm{av}}}^2=2.2\times {10}^{-7}{\mathrm{V}}^2 \end{gathered}$$