For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passes through it. If the torsional constant of suspension wire is $4.0\times10^{-5}\mathrm{N~m~rad^{-1}}$, the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm$^2$) is :
Solution
$\because \theta=\left(\frac{N B A}{K}\right) I$
<br/><br/>
$$
\begin{aligned}
A & =\frac{\theta K}{N B I} \\\\
& =\frac{0.05 \times 4 \times 10^{-5}}{(200) \times(0.01) \times\left(10 \times 10^{-3}\right)} \\\\
& =1 \mathrm{~cm}^{2}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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