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 $NA^{-1}m^{-1} \, is $

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

Answer 1

Answer: (D)

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