Question

In a moving coil galvanometer, the coil having $ 500 $ turns and an average area of $ 4 \,cm^{2} $ carries a current of $ 0.1 \,A $ and is set parallel to a magnetic field with a torque of $ 0.15\, N \,m $ . The strength of the magnetic field is

• A. $ 1\, T $
• B. $ 5T $
• C. $ 7.5 \, T $
• D. $ 50\, T $

Answer 1

Answer: (C)

Here,
$N=500, A=4\,cm^{2}$
$=4\times10^{-4}m^{2}$
$I=0.1\,A$,
$\tau=0.15\,N\,m$
As torque, $\tau=NIAB$
$\therefore B=\frac{\tau}{NIA}$
$=\frac{0.15\,Nm}{\left(500\right)\left(0.1A\right)\left(4\times10^{-4}m^{2}\right)}$
$=7.5\,T$