Question

A solenoid of length $0.4\text{m}$ and diameter $0.6\text{m}$ consists of a single layer of $1000$ turns of fine wire carrying a current of $5.0\text{mA}.$ Find the magnetic field on the axis at the ends of the solenoid.
[Take, ${\mu }_0=4\pi \times {10}^{-7}{\text{Tm A}}^{-1}$]

• A. $2\pi \times {10}^{-6}\text{T}$
• B. $\pi \times {10}^{-6}\text{T}$
• C. $\frac{\pi }{2}\times {10}^{-6}\text{T}$
• D. $\frac{1}{\pi }\times {10}^{-6}\text{T}$

Answer 1

The correct option is A $2\pi \times {10}^{-6}\text{T}$


The magnetic field inside the solenoid at a point is,

$B=\frac{{\mu }_{0}ni}{2}[\cos {\theta }_1-\cos {\theta }_2]......\left(1\right)$

At the end of the solenoid,


From the figure, we get,

$\cos {\theta }_1=\frac{0.4}{\sqrt{(0.3{)}^2+(0.4{)}^2}}=0.8$

$\cos {\theta }_2=\cos {90}^{\circ }=0$

$n=\frac{N}{L}=\frac{1000}{0.4}=2500$

Putting this value in $\left(1\right)$ we get,

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

$=\frac{4\pi \times {10}^{-7}\times 2500\times 5\times {10}^{-3}\times 0.8}{2}$

$\therefore B=2\pi \times {10}^{-6}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(\text{A}\right)$ is the correct answer.

Why this question ? To understand the concept of magnetic field due to a solenoid.