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. Suppose the ends of the coil are connected to a resistance of 100 Ω. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.

Answer 1

Given:-

T = 1 minute

Heat produced in the circuit is calculated using the following relation:-

\[H = \int\limits_0^T i^2 Rdt\]

\[\Rightarrow H =  \int\limits_0^{1  \min} \frac{B^2 A^2 \omega^2}{R^2}\sin\left( \omega t \right)Rdt\]

\[= \frac{B^2 A^2 \omega^2}{2R} . \int\limits_0^{1 \min} \left( 1 - \cos 2\omega t \right)dt\]

\[ = \frac{B^2 A^2 \omega^2}{2R} \left( 1 - \frac{\sin2\omega t}{2\omega} \right)_0^{1 \min} \]

\[ = \frac{B^2 A^2 \omega^2}{2R}\left( 60 - \frac{\sin 2 \times 80 \times 2\pi/60 \times 60}{2 \times 80 \times 2\pi/60} \right)\]

\[ = \frac{60}{2R} \times \pi^2 r^4 \times B^2 \times \left( 80 \times \frac{2\pi}{60} \right)^2 \]

\[ = \frac{60}{200} \times 10 \times \frac{64}{9} \times 10 \times 625 \times {10}^{- 8} \times {10}^{- 4} \]

\[ = \frac{625 \times 6 \times 64}{9 \times 2} \times {10}^{- 11} = 1 . 33 \times {10}^{- 7} J\]