Question

A long straight conductor is bent into shape as shown. If it carries $1A$ and its radius is R, then magnetic field B at the centre of circular coil

• A. $\infty$
• B. Zero
• C. $\frac{{\mu }_{0}l(\pi +1)}{2\pi R}$
• D. $\frac{{\mu }_{0}l(\pi -1)}{2\pi R}$

Answer 1

The correct option is D $\frac{{\mu }_{0}l(\pi -1)}{2\pi R}$

Given : Current flowing $i=1$ $A$

For infinite long straight wire :

Magnetic field at the centre of coil $B_1=\frac{{\mu }_{o}i}{2\pi R}$

$⟹$ $B_1=\frac{{\mu }_o}{2\pi R}$ (into the paper)

For circular loop :

Magnetic field at the centre of coil $B_2=\frac{{\mu }_{o}i}{2R}$

$⟹$ $B_2=\frac{{\mu }_o}{2R}$ (coming out of the paper)

Thus total magnetic field at the centre of coil, $B_T=B_2-B_1$

$\therefore$ $B_T=\frac{{\mu }_o}{2R}-\frac{{\mu }_o}{2\pi R}=\frac{{\mu }_o(\pi -1)}{2\pi R}$ (coming out of the paper)