Question

The wire shown in figure carries a current of 40A. The radius of the circular part of the wire is 2 cm. Find the magnetic field at the centre of the circular part of the wire.

• A. $9.43\times 10^{-4}T$
• B. $9.43\times 10^{-6}T$
• C. $4.93\times 10^{-4}T$
• D. $6\times 10^{-6}T$

Answer 1

Answer: (A)

Magnetic field due to current element $I\overrightarrow{dI}$ at O is given by $dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{r^2}=\frac{{\mu }_0}{4\pi }\frac{Idl}{r^2}$ ($\because$ sin 90${}^{\circ }$ = 1) $\therefore$ magnetic field at O due to the whole circular part AB is B = $\frac{{\mu }_0}{4\pi }\frac{I}{r^2}\underset{0}{\overset{\frac{3}{4}\times 2\pi r}{\int }}$ = $\frac{{\mu }_0}{4\pi }\frac{I}{r^2}\times \frac{3}{4}\times 2\pi r=\frac{{\mu }_0}{4\pi }\left(\frac{3I\pi }{2r}\right)$ B = $\frac{10^{-7}\times 3\times 40\times 3.142}{2\times 2\times 10^{-2}}=9.43\times 10^{-4}$T