Question

A part of a long wire carrying a current I is bent into a circle of radius r as shown in Fig. The net magnetic field at the centre O of the corcular loop is

• A. $\frac{{\mu }_{0}I}{4r}$
• B. $\frac{{\mu }_{0}I}{2r}$
• C. $\frac{{\mu }_{0}I}{2\pi r}(\pi +1)$
• D. $\frac{{\mu }_{0}I}{2\pi r}(\pi -1)$

Answer 1

The correct option is C

$\frac{{\mu }_{0}I}{2\pi r}(\pi +1)$

Magnetic field at point 0 due to straight wire AB is
$B_1=\frac{{\mu }_0}{4\pi }.\frac{2I}{r}=\frac{{\mu }_{0}I}{2\pi r}$
Both these fields are normal to the plane of the loop and directed into the page. Therefore, net field at point 0 is
$B=B_1+B_2=\frac{{\mu }_{0}I}{2\pi r}+\frac{{\mu }_{0}I}{2r}=\frac{{\mu }_{0}I}{2\pi r}(1+\pi )$
Hence the correct choice is (c).