Question

A long solenoid of $100\text{cm}$ length, having $200$ turns, carries $5\text{A}$ current. The magnetic field at one end of the solenoid is
$({\mu }_0=4\pi \times {10}^{-7}\text{Tm/A})$

• A. $6.28\times {10}^{-4}\text{T}$
• B. $2.6\times {10}^{-4}\text{T}$
• C. $12.4\times {10}^{-4}\text{T}$
• D. $9.81\times {10}^{-3}\text{T}$

Answer 1

The correct option is A $6.28\times {10}^{-4}\text{T}$

At end of a long solenoid, magnetic field is given as $B=\frac{{\mu }_{0}ni}{2}$, where $n$ is number of turns per unit length.

$B=\frac{4\pi \times {10}^{-7}\times \frac{200}{1}\times 5}{2}=6.28\times {10}^{-4}\text{T}$