Question

A solenoid of length $50cm$, having $100$ turns carries a current of $2.5A$. Find the magnetic field (B),(a) in the interior of teh solenoid, (b) at one end of the solenoid. Given ${\mu }_0=4\pi \times {10}^{-7}Wb{A}^{-1}{m}^{-1}$.

Answer 1

Given : $l=50cm=0.5m$ $i=2.5A$ $N=100$ turns

So, number of turns per unit length $n=\frac{N}{l}=\frac{100}{0.5}=200turns/m$

(i) : Magnetic field inside the solenoid $B={\mu }_{o}ni=4\pi \times {10}^{-7}\times 200\times 2.5=6.28\times {10}^{-4}T$

(ii) : Magnetic field at the end of solenoid $B_{end}=\frac{{\mu }_{o}ni}{2}=\frac{6.28\times {10}^{-4}}{2}=3.14\times {10}^{-4}T$