Question

A charge q coulomb makes n revolutions in one second in a circular orbit of radius r. The magnetic field at the centre of the orbit in $N{A}^{-1}{m}^{-1}$ is

• A. $\frac{2\pi rn}{q}\times {10}^{-7}$
• B. $\left(\frac{2\pi rn}{q}\right)\times {10}^{-7}$
• C. $\left(\frac{2\pi q}{nr}\right)\times {10}^{-7}$
• D. $\left(\frac{2\pi nq}{r}\right)\times {10}^{-7}$

Answer 1

The correct option is D $\left(\frac{2\pi nq}{r}\right)\times {10}^{-7}$

Magnetic field at the centre of the circular orbit is given by
$B=\left(\frac{{\mu }_{0}I}{2r}\right)$
$B=\left(\frac{2{\mu }_{0}I}{4r\pi }\right)\times \pi$
We get $B=\frac{2{\mu }_{0}I\pi }{4\pi r}$
Thus $B=\frac{2\pi I}{r}\times {10}^{-7}$ $[\because \frac{{\mu }_0}{4\pi }={10}^{-7}]$
$⟹B=\left(\frac{2\pi nq}{r}\right)\times {10}^{-7}$ $[\because I=nq]$