Question

A circular coil of $70$ turns and radius $5\, cm$ carrying a current of $8\, A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.5 \,T$. The field lines make an angle of $30^{\circ}$ with the normal of the -coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is

• A. $33\, N \,m$
• B. $3.3\, N\, m$
• C. $3.3 \times 10^{-2}\, N\, m$
• D. $3.3 \times 10^{-4}\, N\, m$

Answer 1

Answer: (B)

$N=70$,
$r=5\, cm=5 \times 10^{-2}\, m$,
$I=8\, A$
$B=1.5 \, T$,
$\theta = 30^{\circ}$
The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil,
$\therefore \tau= NIAB\, sin\, \theta = NI \pi r^{2}B \, sin\, 30^{\circ}$
$=70\times 8\times 3.14 \times (5 \times 10^{-2})^{2} \times 1.5 \times {\frac {1}{2}}= 3.297\, N\,m$
$\approx 3.3 \, N\, m$