JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 17)
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 :
1.5
2.0
0.5
1.0
Explanation
$\because \theta=\left(\frac{N B A}{K}\right) I$
$$ \begin{aligned} A & =\frac{\theta K}{N B I} \\\\ & =\frac{0.05 \times 4 \times 10^{-5}}{(200) \times(0.01) \times\left(10 \times 10^{-3}\right)} \\\\ & =1 \mathrm{~cm}^{2} \end{aligned} $$
$$ \begin{aligned} A & =\frac{\theta K}{N B I} \\\\ & =\frac{0.05 \times 4 \times 10^{-5}}{(200) \times(0.01) \times\left(10 \times 10^{-3}\right)} \\\\ & =1 \mathrm{~cm}^{2} \end{aligned} $$
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