Question

A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be
A nBAω
B $\dfrac{3}{2}$nBAω
C 3nBAω
D $\dfrac{1}{2}$nBAω

Answer 1

Hint: Electromotive Force or EMF is said to be induced when the flux linking with a conductor or coil changes. The induced emf in a coil is equal to the rate of change of flux linkage $\varepsilon = - \dfrac{{d\varphi }}{{dt}}$.
Where, ε is the electromotive force
Φ is the magnetic flux
n is the number of turns.

Complete step by step answer:
Given, the cross sectional area of the coil =A
Number of turns in the coil= n
Uniform magnetic field=B
Angular velocity=ω
We can write the flux linked to the coil as$\varphi = $nBA cos(ωt).
According to Faraday's law due to the flux change an EMF will be induced in the coil. The rate of EMF in the coil is equal to the rate of change of flux linkage, $\varepsilon = - \dfrac{{d\varphi }}{{dt}}$.
Substituting the value of flux we get the EMF in the coil $\varepsilon = - \dfrac{{d\left[ {nBA\cos (\omega t)} \right]}}{{dt}}$.
Since n, B, A are constant, we get coil $\varepsilon = - nBA\dfrac{{d\left[ {\cos (\omega t)} \right]}}{{dt}}$
 $
\varepsilon = - nBA( - \omega \sin (\omega t)) \\
\therefore \varepsilon = \omega nBA\sin (\omega t) \\
$
For this EMF to be maximum sin should be equal to 1 since the maximum value of sin is 1.

Therefore the maximum EMF induced in the coil is $\varepsilon = nBA\omega $.

Note:
The negative sign indicates that the direction of the induced emf and change in the direction of magnetic fields have opposite signs. EMF can be induced in a coil due to the following reasons:
Increase in the number of turns in the coil increases the induced emf.
Increasing the magnetic field strength increases the induced emf.
Increasing the speed of the relative motion between the coil and the magnet, results in the increased emf.