Question

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 $180°$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?

• A. $\frac{QR}{nA}$
• B. $\frac{2QR}{nA}$
• C. $\frac{Qn}{2RA}$
• D. $\frac{QR}{2nA}$

Answer 1

The correct option is D $\frac{QR}{2nA}$

From Lenz's law we know that
$i=\frac{-N\frac{d\varphi }{dt}}{R}$
$\int idt=-\frac{N}{R}({\varphi }_2-{\varphi }_1)$
where ${\varphi }_2=BAcos{\theta }_2$ and ${\varphi }_1=BAcos{\theta }_1$
As we know $Q=\int idt$
We get
Induced charge,
$Q=-\frac{nBA}{R}(\cos {\theta }_2-\cos {\theta }_1)$
$=\frac{-nBA}{R}\left(\cos 180°-\cos 0°\right)$
$\Rightarrow B=\frac{QR}{2nA}$ .