The coil in a moving coil Galvanometer experiences torque proport… - Vidyadip

MCQ

The coil in a moving coil Galvanometer experiences torque proportional to current passed through it. If a steady current $i$ is passed through it the deflection of the coil is found to be $90^o$ . Now the steady current is switched off and a charge $Q$ is suddenly passed through the coil. If the coil has $N$ turns of area $A$ each and its moment of inertia is $I$ about the axis it is going to rotate then the maximum angle through which it deflects upon passing $Q$ is

A
$Q\sqrt {\frac{{\pi NAB}}{{2iI}}} $
B
$\frac{{\pi Q}}{2}\sqrt {\frac{{NAB}}{{iI}}} $
C
$\pi Q\sqrt {\frac{{NAB}}{{2iI}}} $
D
$Q\sqrt {\frac{{\pi NAB}}{{iI}}} $

✓

Answer

$\tau=\mathrm{MB}=\mathrm{C} \theta$

$ \Rightarrow {\rm{N}}i{\rm{AB}} = {\rm{C}}\frac{\pi }{2} \Rightarrow {\rm{C}} = \frac{{2{\rm{N}}i{\rm{AB}}}}{\pi }$ ........$(1)$

When $Q$ is passed,

$\int \tau dt = \int {Ni} ABdt$

$\mathrm{I} \omega=\mathrm{N} \mathrm{A} \mathrm{B} \mathrm{Q}$ ........$(2)$

Also, maximum deflection happens when entire $\frac{1}{2} \mathrm{I} \omega^{2}$ converts to $\frac{1}{2} \mathrm{C} \theta_{\mathrm{max}}^{2}$

So ${\theta _{\max }} = \sqrt {\frac{{{\rm{I}}{\omega ^2}}}{{\rm{C}}}} = \frac{1}{{\sqrt {{\rm{IC}}} }}{\rm{I}}\omega = \frac{{{\rm{NABQ}}}}{{\sqrt {{\rm{IC}}} }}$ (from $(1)$)

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