Question

A current carrying circular loop is placed in $x-y$ plane, with the centre at origin. Half of the loop is now bent so that it is in $y-z$ plane. If initially, the magnitude of magnetic field at origin is $B$, then what is the magnitude of magnetic field at origin after bending?

• A. $\frac{B}{\sqrt{2}}$
• B. $\sqrt{2}B$
• C. $B^2$
• D. $3B$

Answer 1

The correct option is A $\frac{B}{\sqrt{2}}$

Before bending,

${\overrightarrow{B}}_O=B\hat{k}$

After bending,


${\overrightarrow{B}}_O=\frac{B}{2}\hat{i}+\frac{B}{2}\hat{k}$

So, $B_O=\sqrt{{\left(\frac{B}{2}\right)}^2+{\left(\frac{B}{2}\right)}^2}=\frac{B}{\sqrt{2}}$

Why this question? This question has been asked to make you understand that magnetic field due to semi-circular wire is half of magnetic field due to a circular wire at centre.