Question

The magnetic field due to a current carrying circular loop of radius $10\text{cm}$ at a point on the axis, at a distance of $10\text{cm}$ from the centre, is $B$. What will be its value at the centre of the loop?

• A. $\sqrt{8}B$
• B. $4B$
• C. $2B$
• D. $B$

Answer 1

The correct option is A $\sqrt{8}B$


Let $P$ is the point on the axis of the circular loop.

Given, $R=10\text{cm},x=10\text{cm}$

At point $P$,

$B=\frac{{\mu }_{o}I{R}^2}{2(R^2+x^2{)}^{3/2}}$

$\Rightarrow B=\frac{{\mu }_{o}I}{2^{5/2}R}[\because x=R]$

Also, magnetic field at the centre of the loop is,

$B_O=\frac{{\mu }_{o}I}{2R}=2^{3/2}B=\sqrt{8}B$

Hence, option $\left(A\right)$ is the correct answer.