Question

An ideal solenoid having $n=500\text{turns/m}$ and current $i=0.5\text{A}$. What will be the magnetic field at a point $\text{A}$ situated at the edge of the solenoid just below the end of the solenoid as shown in figure

• A. $1.57\times {10}^4\text{T}$
• B. $0.57\times {10}^{-4}\text{T}$
• C. $1.57\times {10}^{-4}\text{T}$
• D. $15.7\times {10}^{-4}\text{T}$

Answer 1

The correct option is C $1.57\times {10}^{-4}\text{T}$

Given:
$n=500;i=0.5\text{A}$

Assuming, radius of solenoid is very small as compare to length of the same.

So, the magnetic field at the end of the solenoid is given by

$B=\frac{{\mu }_{0}ni}{2}[\cos {\theta }_1-\cos {\theta }_2]$

So, at point $\text{A}$, ${\theta }_1=0^{\circ };{\theta }_2={90}^{\circ }$

$B=\frac{4\pi \times {10}^{-7}\times 500\times 0.5}{2}[\cos 0^{\circ }-\cos {90}^{\circ }]$

$B=4\pi \times {10}^{-7}\times 125$

$\therefore B=1.57\times {10}^{-4}\text{T}$

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