Question

A circular coil of $100$ turns, radius $10\, cm$ carries a current of $5 \,A$. It is suspended vertically in a uniform horizontal magnetic field of $0.5\, T$ and the field lines make an angle of $ 60^{\circ}$ with the plane of the coil. The magnitude of the torque that must be applied on it to prevent it from turning is

• A. $2.93 \,N\,m$
• B. $3.43 \,N\,m$
• C. $3.93 \,N\,m$
• D. $4.93 \,N\,m$

Answer 1

Answer: (C)

Here,
$N = 100$,
$r = 10 \,cm = 0.10 \,m$,
$I = 5 \,A$,
$B = 0.5\, T$,
$\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$
Area of the coil,
$A = \pi r^{2} = 3.14 \times \left(0.1\right)^{2}$
$\tau = NIBAsin\theta$
$ = 100 \times 5 \times 0.5 \times 3.14 \times \left(0.1\right)^{2} \times sin \,30^{\circ}$
$= 3.931 \,N\,m$