Question

The number of turns in the coil of an AC generator is 5000 and the area of the coil is 0.25 ${m^2}$. The coil is rotated at the rate of 100 cycles/sec in the magnetic field of 0.2 $W/{m^2}$. Then the peak value will be:
(A) 786 kV
(B) 178 kV
(C) 157 kV
(D) 123 kV

Answer 1

Hint In this question, we are given the number of coils, magnetic field, and the area of each coil is given. The emf of a coil rotating in a magnetic field such that the face of the coil cuts the magnetic field is directly proportional to the intensity of the magnetic field, several turns angular speed of the coil and $\sin (\omega t)$ .

Complete step by step answer:
We are given the number of turns in the coil of the AC generator, N=5000.
Area of the cross section of the coil is given to be, $A = 0.25{m^2}$.
The coil is rotated at the rate of 100 cycles/ sec. $\Rightarrow \omega = 2\pi \times 100 = 628{\text{cycles/sec}}$
The magnetic field across the coil is, $B = 0.2{\text{ W/}}{{\text{m}}^{\text{2}}}{\text{ or T}}$ .
Peak value of EMF will be: ${\varepsilon _0} = \omega NBA$
Substituting all the given values in the above equation,
$\Rightarrow {\varepsilon _0} = \omega NBA \\\ \Rightarrow {\varepsilon _0} = 628 \times 5000 \times 0.2 \times 0.25 \\\ \Rightarrow {\varepsilon _0} = 157000V \\\ \Rightarrow {\varepsilon _0} = 157kV \\\$

Therefore, option (C) is correct.

Note Generally, the formula for the EMF induced in the coil of an AC generator is written as $\varepsilon = \omega NBA\sin \omega t$. The value of $\sin (\omega t)$ varies from +1 to -1. For the emf to be at its maximum value, $\sin (\omega t)$ must be equal to +1 and for the minimum value of the emf induced, it should be $\sin (\omega t)$. Keep in mind that the minimum value here means the maximum value but in the opposite direction.