Question

A moving coil galvanometer has a coil of area $A$ and number of turns $N$. A magnetic field $B$ is applied on it. Moment of inertia of the coil is $I$ about the axis of rotation.
$\left(a\right)$ Find the torsion constant, $C$ of the spring, if a current $i_0$ produces a deflection of $\frac{\pi }{2}$ in the coil.
$\left(b\right)$ If a charge $Q$ is passed instantaneously through the coil, find the maximum angular deflection, ${\theta }_0$ in it.
$(\text{Maximum torsional energy}=\frac{1}{2}C{\theta }_0^2)$

• A. $C=\frac{N{i}_{0}AB}{\pi }$
• B. $C=\frac{2N{i}_{0}AB}{\pi }$
• C. ${\theta }_0=Q\sqrt{\frac{\pi NAB}{2i_{0}I}}$
• D. ${\theta }_0=Q\sqrt{\frac{\pi NAB}{i_{0}I}}$

Answer 1

Answer: (B), (C)

${\theta }_0=Q\sqrt{\frac{\pi NAB}{2i_{0}I}}$

$\left(a\right)$ Torque acting on the coil when angle of defection is $\frac{\pi }{2}$,

$\tau =N{i}_{0}AB\sin {90}^{\circ }=N{i}_{0}AB.....\left(1\right)$

The torsion constant of a coil is,

$C=\frac{\tau }{\theta }$

$C=\frac{N{i}_{0}AB}{\pi /2}=\frac{2N{i}_{0}AB}{\pi }.....\left(2\right)$

So, option $\left(B\right)$ is the correct option.

$\left(b\right).$ If ${\theta }_0$ is the maximum deflection, then by conservation of energy

$\frac{1}{2}C{\theta }_0^2=\frac{1}{2}I{\omega }^2$

$\Rightarrow {\theta }_0=\sqrt{\frac{I}{C}}\times \omega .........\left(3\right)$

If $L$ is the angular momentum of the coil, then $\tau =\frac{dL}{dt}$, and equation $\left(1\right)$ can be written as

$\frac{dL}{dt}=N\frac{dQ}{dt}AB$

$\Rightarrow dL=NABdQ$

On integrating, we get

${\int }_0^{L}dL=NAB{\int }_0^{Q}dQ$

$\Rightarrow L=NABQ$

$\Rightarrow I\omega =NABQ[\because L=I\omega ]$

$\Rightarrow \omega =\frac{NABQ}{I}.......\left(4\right)$

Using equations $\left(2\right),\left(3\right)$ and $\left(4\right)$, we get

${\theta }_0=\sqrt{\frac{I}{\frac{2N{i}_{0}AB}{\pi }}}\times \frac{NABQ}{I}$

$\therefore {\theta }_0=Q\sqrt{\frac{\pi NAB}{2i_{0}I}}$

So, option $\left(C\right)$ is also correct.