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. (3/2)nBAω
• C. 3nBAω
• D. (1/2)nBAω

Answer 1

The correct option is A

nBAω

Step 1: Given

Given, a coil with the area of cross-section A and n turn is given which is to be placed in a uniform magnetic field B. later coil is rotated with an angular velocity ω then the maximum e.m.f. ${\epsilon }_0$ induced in the coil is required.

Diagram representing the given condition-

Step 2: Calculate the magnetic flux $\varphi$

The magnetic flux linking to the coil, $$\begin{gathered} \varphi =n\overrightarrow{B}.\overrightarrow{A} \\ \varphi =nBA\cos \theta \end{gathered}$$

now, angular velocity $$\begin{gathered} \omega =\frac{\theta }{t} \\ \theta =\omega t \end{gathered}$$

here, $\theta$ is the angle of rotation of coil with time t.

then, $\varphi =nBA\cos \omega t$

Step 3: Calculate the emf $\epsilon$ and Maximum emf ${\epsilon }_0$

therefore, the emf induced in the coil, $$\begin{gathered} \epsilon =-\left(\frac{d\varphi }{dt}\right) \\ \epsilon =-\frac{d\left(nBA\cos \omega t\right)}{dt} \\ \epsilon =-nBA\left(-\sin \omega t\right)\omega \\ \epsilon =nBA\omega \sin \omega t \end{gathered}$$

now, the emf will be maximum at $\sin \omega t=1$

$$\begin{gathered} \sin \omega t=1 \\ \omega t=90° \\ \theta =90° \end{gathered}$$

so, the maximum emf ,

$$\begin{gathered} {\epsilon }_0=nBA\omega .1 \\ {\epsilon }_0=nBA\omega \end{gathered}$$

Thus, the maximum emf ${\epsilon }_0$ induced in the coil is $nBA\omega$.

Hence, option A is correct.