Question

A circular coil of radius $2.5\text{cm}$ carries current of $2\text{A}$. If the coil has $50$ turns. Find the magnetic field at the center of the coil ? <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->

• A. $\frac{\pi }{6250}\text{T}$
• B. $\frac{\pi }{625}\text{T}$
• C. $\frac{\pi }{125}\text{T}$
• D. $\frac{\pi }{1250}\text{T}$

Answer 1

The correct option is D $\frac{\pi }{1250}\text{T}$

Given,

$I=2\text{A},n=50\text{turns}$ and

$R=2.5\times {10}^{-2}\text{m}$

The magnetic field at the centre of the coil is,

$B=\frac{{\mu }_{0}nI}{2R}$

$=\frac{4\pi \times {10}^{-7}\times 2\times 50}{2\times 2.5\times {10}^{-2}}$

$\therefore B=\frac{\pi }{1250}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (d) is the correct answer.