Question

A rectangular coil of $300$ turns has an average area of $25\;cm \times 10\;cm$. The coil rotates with a speed of $50\;cps\;$ in the uniform magnetic field of strength $4 \times {10^{ - 2}}T\;$ about an axis perpendicular to the field. The peak value of the induced emf is (in volt):
A. $300\pi$
B. $3000\pi$
C. $3\pi$
D. $30\pi$

Answer 1

Change in the magnetic flux induces emf which opposes that change. This process is known as induction. Faraday’s law states that the magnitude of the electromotive force (emf) is directly proportional to the rate of change of the magnetic field. How dense the field lines of a magnetic field are within a given height and the strength of the magnetic field is called the magnetic flux density.

Complete step by step solution:
Given that the rectangular coil has $300$turns. That is $N = 300$
The area of the rectangular coil $A = 25 \times 10c{m^2}$
Converting$c{m^2}$to${m^2}$we get, $A = 250 \times {10^{ - 4}}{m^2}$
The speed with which the coil rotates that is the frequency of the coil is given as $f = 50$
Magnetic field strength is given as $B = 4 \times {10^{ - 2}}T\;$
The induced emf on the given coil is given by the formula,
$e = \omega NBA\sin (\omega t)$ ………… (1)
Here we are asked to find the induced peak voltage. When the longer side moves perpendicular to the magnetic field peak voltage will be induced. $\sin (\omega t)$ should be equal to$1$for the voltage to be in its peak.
Therefore equation (1) becomes,
$e = \omega NBA$ ………..(2)
We know that $\omega = 2\pi f$
Substituting all values in equation (2) we get,
$e = 2\pi (50) \times 300 \times 4 \times {10^{ - 2}} \times \;250 \times {10^{ - 4}}$
Solving the above expression we will get
$e = 30\pi$
Therefore the peak induced voltage is given as $30\pi$volts.
Hence the correct option is D.

Note:
The fact that the peak emf$em{f_{0\;}} = \;NAB\omega$makes good sense. We can see that the induced voltage is directly proportional to the number of coils, area, and magnetic field strength. This means that the greater the number of coils, the larger area, and the stronger field give a greater output voltage. The interesting fact is the faster the generator is spun (greater$\omega$), the greater the emf.