Search any question & find its solution
Question:
Answered & Verified by Expert
A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is $n$ and the cross-sectional area of the coil is $A$. When the coil turns through $180^{\circ}$ about its diameter, the charge flowing through the coil is Q. The total resistance of the circuit is $R$. What is the magnitude of the magnetic induction?
Options:
Solution:
1860 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{QR}}{2 \mathrm{nA}}$
Induced charge
$$
\begin{aligned}
Q &=-\frac{n B A}{R}\left(\cos \theta_{2}-\cos \theta_{1}\right) \\
&=-\frac{n B A}{R}\left(\cos 180^{\circ}-\cos 0^{\circ}\right) \\
\Rightarrow B &=\frac{Q R}{2 n A}
\end{aligned}
$$
$$
\begin{aligned}
Q &=-\frac{n B A}{R}\left(\cos \theta_{2}-\cos \theta_{1}\right) \\
&=-\frac{n B A}{R}\left(\cos 180^{\circ}-\cos 0^{\circ}\right) \\
\Rightarrow B &=\frac{Q R}{2 n A}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.