QJEEAdv18Phcae2: JEE Advanced 2018 Physics - Class 12 JEEAdv18Phcae2 Solution: A moving coil galvanometer has turns and each turn has an area…

The magnetic moment of a coil having n turns, area A and current i is given by $\overrightarrow \mu = ni\overrightarrow A$ . The torque on this coil when placed in a magnetic field $\overrightarrow B$ is given by $\overrightarrow \tau = \overrightarrow \mu \times \overrightarrow B$ . In a moving coil galvanometer, cylindrical magnets ensure that $\overrightarrow A$ and $\overrightarrow B$ are perpendicular to each other. Thus, deflection torque on the coil is $\tau$ _d = niAB. The restoring torque on a coil (at deflection angle $\theta$ ) due to the wire of torsional constant k is $\tau$ _r = k $\theta$ . In equilibrium, deflection torque is balanced by the restoring torque i.e., $\tau$ _d = $\tau$ _r, which gives

$i = {k \over {nAB}}\theta$ .

The maximum deflection current of the galvanometer (i_g) occurs at full scale deflection $\theta$ _max i.e.,

${i_g} = {{k{\theta _{\max }}} \over {nAB}} = {{({{10}^{ - 4}})(0.2)} \over {(50)(2 \times {{10}^{ - 4}})(0.02)}} = 0.01$ A.

A galvanometer of resistance G is converted to an ammeter by connecting a small shunt resistance S in parallel.

Kirchhoff's loop law gives

${i_g}G - (i - {i_g})S = 0$ , $\Rightarrow S = {i_g}G/(i - {i_g})$ .

The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter (i_max). Substitute the values to get

$S = {{{i_g}G} \over {{i_{\max }} - {i_g}}} = {{(0.1)(50)} \over {1 - 0.1}} = {5 \over {0.9}} = 5.55\Omega$ .