Question

In a region of uniform magnetic induction B =${10}^{-2}$ , a circular coil of radius 30 cm and resistance ${\pi }^2$ ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm. Find the amplitude of the alternating current induced in the coil?

Answer 1

Given,

$n=1,B={10}^{-2}T$

$A=p(0.3{)}^2,m^2,R=p^2$

$f=\frac{200}{60}$ and $\omega =2\pi \times \frac{200}{60}$

So, the alternating current induced in the coil is:

$I=\frac{{\epsilon }_0}{R}=\frac{nBA\omega }{R}$

$=\frac{1\times {10}^{-2}{p}^2(0.3{)}^2{m}^2\times 2\pi \times 20}{p^2\times 60}$

$=6\times {10}^{-3}A$

$=6mA$