Question

A circular coil of 950 turns and a radius of 0.060 m is rotating in a uniform magnetic field of 8.56 x 10-5 T. The coil rotates from an angle of 0° to 45° in 0.10 s. Calculate the magnitude of the induced emf in the coil. Please answer as soon as possible.

Answer 1

Answer:  $2.69 \times 10^{-3} V$

Explanation:

A circular coil has number of turns N = 950 and it's radius r = 0.06 m

The coil is rotating in a uniform magnetic field  $B = 8.56 \times 10^{-5} T$.

The coil rotates from an angle of 0° to 45° in 0.10 s

We have to calculate the induced emf in the coil,

$\epsilon_{induced} = -N\frac{\Delta \phi }{\Delta t} = -N\frac{\Delta (B A cos(\phi))} {\Delta t} = -NBA\frac{\Delta (cos(\phi))}{\Delta t}$

Here, A is the area of that coil and  $\phi$  is the angle of the coil with the field at time t. In question it is said that, at $t_1 = 0$ s the angle is  $\phi_1 = 0^{0}$ and at time $t_2 = 0.1$ s the angle becomes  $\phi_2 = 45^{0}$.  

So, $\epsilon_{induced} = -NBA \frac{(cos \phi_2- cos \phi_1)}{t_2 - t_1}$

                  $= - 950 \times 8.56 \times 10^{-5} \times \pi \times (0.06)^{2} \times \frac{cos45^{0}-cos0^{0} }{0.1} V$

                   $= - 9.19 \times 10^{-3} \times (- 0.29289) V\\ = 2.69 \times 10^{-3} V$