Question

The magnetic field at $P$ on the axis of a solenoid having $100$ turn/ $m$ and carrying a current of $5\, A$ is

• A. $250\, \mu_{0}$
• B. $500 \,\sqrt{2} \mu_0$
• C. $500 \,\mu_0$
• D. $250 \,\sqrt{2} \mu_0$

Answer 1

Answer: (D)

$250 \,\sqrt{2} \mu_0$

Answer 2

The magnetic field at $P$ is $B=\frac{\mu_{0} n I}{2}(\cos \theta+\cos \theta)$ where $n$ is number of turns, $I$ the current. Given, $n=100, I=5 A$ and $\theta=45^{\circ}$ $\therefore B=\frac{\mu_{0} \times 100 \times 5}{2} \times \frac{2}{\sqrt{2}}$ $\Rightarrow B =250 \sqrt{2} \,\mu_{0}$