Question

A solenoid of length $0.4$ m and having $500$ turns of wire carries a current $3$ A. A thin coil having $10$ turns of wire and radius $0.1$ m carries a current $0.4$ A. The torque produced to hold the coil in middle of solenoid with its axis perpendicular to the axis of the solenoid is $(\mu_0=4\pi\times10^{-7}$ SI units, $\pi^2=10$, $\sin90^\circ=1)$

• A. $3\times10^{-6}\ \text{Nm}$
• B. $12\times10^{-6}\ \text{Nm}$
• C. $6\times10^{-4}\ \text{Nm}$
• D. $24\times10^{-6}\ \text{Nm}$

Hint:

Physics Tip : Torque on coil in magnetic field: $\tau=N I A B \sin\theta$.

Answer 1

The Correct Option is C

Step 1: Magnetic field inside solenoid. Field inside long solenoid: $$ B=\mu_0 n I $$ where $$ n=\frac{N}{L}=\frac{500}{0.4}=1250\ \text{turns/m} $$ Thus, $$ B=4\pi\times10^{-7}\times1250\times3 $$

Step 2: Magnetic moment of coil. For circular coil: $$ m=NIA $$ Given:
• $N=10$
• $I=0.4$ A
• $r=0.1$ m Area: $$ A=\pi r^2=\pi(0.1)^2=0.01\pi $$ So, $$ m=10\times0.4\times0.01\pi=0.04\pi $$

Step 3: Torque on magnetic dipole. $$ \tau=mB\sin\theta $$ Axis of coil is perpendicular to solenoid axis: $$ \theta=90^\circ,\quad \sin90^\circ=1 $$ Hence, $$ \tau=mB $$

Step 4: Substitute values. $$ \tau=(0.04\pi)\left(4\pi\times10^{-7}\times1250\times3\right) $$ Using $\pi^2=10$: $$ \tau=6\times10^{-4}\ \text{Nm} $$

Step 5: Conclusion. $$ \therefore \text{Correct option is (C).} $$