Question

A circular loop of area $0.01 \,m ^{2}$ carrying a current of $10\, A$, is held perpendicular to a magnetic field of intensity $0.1\, T$. The torque acting on the loop is

• A. 0.001 N m
• B. 0.8 N m
• C. zero
• D. 0.01 N m

Answer 1

Answer: (C)

Area $( A )=0.01 \,m _{2}$; Current $( I )=10 A$;
Angle $(\phi)=90^{\circ}$ and magnetic field $(B)=0.1 T$.
Therefore acutal angle $\theta=\left(90^{\circ}-\phi\right)=\left(90^{\circ}-90^{\circ}\right)=0^{\circ}$.
And torque acting on the loop $(\tau)=I A B \sin \theta$
$=10 \times 0.01 \times 0.1 \times \sin 0^{\circ}=0$.