Question

A long solenoid of $50\text{cm}$ length having $100$ turns carries a current of $2.5\text{A}$. The magnetic field at the centre of the solenoid is :
$({\mu }_o=4\pi \times {10}^{-7}{\text{Tm A}}^{-1})$

• A. $6.28\times {10}^{-5}\text{T}$
• B. $3.14\times {10}^{-5}\text{T}$
• C. $6.28\times {10}^{-4}\text{T}$
• D. $3.14\times {10}^{-4}\text{T}$

Answer 1

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

Given:
Length of the solenoid, $l=50\text{cm}$
Current, $i=2.5\text{A}$
Nmber of turns, $N=100\text{turns}$

So, number of turns per unit length $n=\frac{N}{l}=\frac{100}{0.5}=200\text{turns/m}$
Magnetic field inside the solenoid: $B={\mu }_{o}ni=4\pi \times {10}^{-7}\times 200\times 2.5$
$\therefore B=6.28\times {10}^{-4}\text{T}$