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 (e) is given by,

\[e = \frac{d\phi}{dt}\]

\[ \Rightarrow e = \frac{dB . A\cos\theta}{dt}\]

\[ \Rightarrow e =  - BA\sin\theta\frac{d\theta}{dt}\]

`rArr e = −BAω sinθ ............((d theta)/(dt)=omega=" the rate of change of angle between the arc vector and B")`

(a) For maximum emf, sinθ = 1

∴ e = BAω

`rArr e=0.010xx25xx10^-4xx80xx(2pixxpi)/60`

⇒ 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

\[{e_{av}}^2  = \frac{\int_0^T B^2 A^2 \omega^2 \sin^2 \omega t  dt}{\int_0^T dt}\]

\[ \Rightarrow  {e_{av}}^2  = \frac{B^2 A^2 \omega^2 \int_0^T \sin^2 \omega t  dt}{\int_0^T dt}\]

\[ \Rightarrow  {e_{av}}^2  = \frac{B^2 A^2 \omega^2 \int_0^T \left( 1 - \cos2\omega t \right)  dt}{2T}\]

\[ \Rightarrow  {e_{av}}^2  = \frac{B^2 A^2 \omega^2}{2T} \left[ t - \frac{\sin2\omega t}{2\omega} \right]_0^T \]

\[ \Rightarrow  {e_{av}}^2  = \frac{B^2 A^2 \omega^2}{2T}\left[ T - \frac{\sin4\pi - \sin  0}{2\omega} \right] = \frac{B^2 A^2 \omega^2}{2}\]

\[ \Rightarrow  {e_{av}}^2  = \frac{(6 . 66 \times {10}^{- 4} )^2}{2} = 22 . 1778 \times  {10}^{- 8}    V^2...............\left[ \because BA\omega = 6 . 66 \times {10}^{- 4} V \right]\]

\[ \Rightarrow  {e_{av}}^2    = 2 . 2 \times  {10}^{- 7}    V^2\]