Question

A rectangular coil of wire of $500$ turns of area $10\times 5c{m}^2$ carries a current of $2A$ in a magnetic field of induction $2\times {10}^{-3}T$ . If the plane of the coil is parallel to the field. The torque on the coil is (in$Nm$):

• A. $0.1$
• B. $0.01$
• C. $0.001$
• D. $1$

Answer 1

Answer: (B)

$0.01$

The torque on the rectangular coil due to presence of magnetic field is given,

$\tau =NIAB\sin \theta$

where number of turns $N=500$,

Current in the coil $I=2A$,

Area of the coil $A=(10\times 5){10}^{-4}{m}^2$,

Magnetic field $B=2\times {10}^{-3}T$,

Angle between area and magnetic field vector is $\theta$

As area vector is always normal to plane and given the plane is parallel to field, so the angle between area and field is ${90}^o$.

So, $\tau =500\times 2\times (10\times 5)\times {10}^{-4}\times (2\times {10}^{-3})sin90=0.01Nm$