Question

The magnetic field at the centre of circular loop in the circuit carrying current $I$ shown in the figure is :

• A. $\frac{{\mu }_0}{4\pi }\frac{2I}{r}(1+\pi )$
• B. $\frac{{\mu }_0}{4\pi }\frac{2I}{r}(\pi -1)$
• C. $\frac{{\mu }_0}{4\pi }\frac{2I}{r}$
• D. $\frac{{\mu }_0}{4\pi }\frac{I}{r}(\pi +1)$

Answer 1

The correct option is B $\frac{{\mu }_0}{4\pi }\frac{2I}{r}(\pi -1)$

Magnetic field due to circular loop is
$B_1=\frac{{\mu }_{0}I}{2r}=\frac{{\mu }_{0}I}{4\pi r}2\pi$
field due to straight wire is
$B_2=\frac{{\mu }_{0}I}{2\pi r}=\frac{{\mu }_{0}I}{4\pi r}2$
$\because$ net field $=B_1-B_2=\frac{{\mu }_{0}I}{4\pi r}(2\pi -2)$
$=\frac{{\mu }_0}{4\pi }\times \frac{2I}{r}(\pi -1)$