Question

A solenoid of length $0.4m$ and diameter $0.6m$ consists of a single layer of 1000 turns of fine wire carrying a current of $5\times {10}^{-3}A$. Calculate the magnetic field on the axis at the middle and at the end of the solenoid :

• A. $8.7\times {10}^{-6}T,6.28\times {10}^{-6}T$
• B. $6.28\times {10}^{-6}T,8.7\times {10}^{-6}T$
• C. $5.7\times {10}^{-6}T,6.28\times {10}^{-6}T$
• D. $8.7\times {10}^{-6}T,8.28\times {10}^{-6}T$

Answer 1

The correct option is A $8.7\times {10}^{-6}T,6.28\times {10}^{-6}T$

At the center of the solenoid:
In case of solenoid the field at a point on the axis magnetic field is given by
$B=\frac{{\mu }_{0}ni}{2}(\sin \alpha +\sin \beta )$
For a point at the center of the solenoid,
$\alpha =\beta$
$\therefore B={\mu }_{o}ni\sin \alpha$
$\therefore B=4\pi \times {10}^{-7}\times \frac{1000}{0.4}\times (5\times {10}^{-3})\times \left(\frac{0.2}{\sqrt{({0.3}^2+{0.2}^2)}}\right)$
$B=5\pi \times {10}^{-6}\times \left(\frac{0.2}{\sqrt{0.04+0.09}}\right)=5\pi \times {10}^{-6}\times 0.555$
$\therefore B=2.775\times \frac{22}{7}\times {10}^{-6}=8.73\times {10}^{-6}Tesla$
At the end of the solenoid:
$\alpha =0^0$
$\sin \beta =\frac{0.4}{\sqrt{{0.4}^2+{0.3}^2}}=\frac{0.4}{0.5}=0.8$
$B=\frac{{\mu }_{0}ni}{2}(\sin \alpha +\sin \beta )$
$\therefore B=\frac{4\pi \times {10}^{-7}\times \frac{1000}{0.4}\times (5\times {10}^{-3})}{2}\times (0+0.8)=2\pi \times {10}^{-6}=6.28\times {10}^{-6}Tesla$