Question

A solenoid of $100$ turns and area of cross-section $1{\text{cm}}^2$ carries a current of $5\text{A}$. It acts as a bar magnet. Find the magnetic field at a point $P$, situated at a distance of $0.5\text{m}$ on its axis.

• A. $8\times {10}^{-8}\text{T}$
• B. $8\times {10}^{-7}\text{T}$
• C. $8\times {10}^{-6}\text{T}$
• D. $8\times {10}^{-5}\text{T}$

Answer 1

The correct option is A $8\times {10}^{-8}\text{T}$

Magnetic moment,

$M=NIA=100\times 5\times {10}^{-4}=0.05{\text{A-m}}^2$

The magnetic field due to a solenoid acting as a bar magnet, at a point (outside)on its axis, is given by,

$B=\frac{{\mu }_0}{4\pi }\frac{2M}{r^3}$

$\Rightarrow B={10}^{-7}\times \frac{2\times 0.05}{{0.5}^3}=8\times {10}^{-8}\text{T}$

Hence, option $\left(A\right)$ is the correct answer.