Question

A solenoid of $100$ turns and having current of $5\text{A}$ acts as a bar magnet having area of cross-section $0.1{\text{cm}}^2$. Find the magnetic field at a point $P$ situated at distance $0.5\text{m}$ from the centre on its axis

• A. $8\times {10}^{-10}\text{T}$
• B. $12.5\times {10}^{-9}\text{T}$
• C. $8\times {10}^{-9}\text{T}$
• D. $0.5\times {10}^{-9}\text{T}$

Answer 1

The correct option is C $8\times {10}^{-9}\text{T}$

The magnetic field due to a solenoid acting as a bar magnet at any point is given by,

$B_a=\frac{{\mu }_0}{4\pi }\frac{M}{r^3}\sqrt{3{\cos }^2\theta +1}$


At axial point $\theta =0^{\circ }$

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

Magnetic moment,

$M=NIA$

$=100\times 5\times 0.1\times {10}^{-4}$

$=5\times {10}^{-3}{\text{Am}}^2$

$\Rightarrow B_a={10}^{-7}\times \frac{2\times 5\times {10}^{-3}}{(0.5{)}^3}=8\times {10}^{-9}\text{T}$

Hence, option $\left(C\right)$ is correct.