Question

A point charge $Q(=3\times 10^{-12}C)$ rotates uniformly in a vertical circle of radius $R(=1mm).$ The axis of the circle is aligned along the magnetic axis of the earth. At what value of the angular speed go, the effective magnetic field at the centre of the circle will be reduced to zero? (Horizontal component of earth’s magnetic field is $30\mu T$)

• A. $10^{11}$ rad/s
• B. $10^9$ rad/s
• C. $10^{13}$ rad/s
• D. $10^7$ rad/s

Answer 1

Answer: (A)

Net magnetic field at centre of loop is zero, when magnetic field due to rotating charge is exactly equal to the horizontal component of earth’s magnetic field.
i.e. $B_q=B_H\Rightarrow \frac{{\mu }_{0}i}{2R}=B_H$
$\Rightarrow \frac{{\mu }_0\left(\frac{q}{T}\right)}{2R}=B_H$
$\Rightarrow \frac{{\mu }_{0}q\omega }{2R\cdot 2\pi }=B_H$
$\Rightarrow \frac{{\mu }_{0}q\omega }{4\pi R}=B_H$
Here $,\frac{{\mu }_0}{4\pi }=10^{-7}Tm{A}^{-1},q=3\times 10^{-12}C$
$R=1mm=10^{-3}m,B_H=30\times 10^{-6}T$
$\Rightarrow \frac{10^{-7}\times 3\times 10^{-12}\times \omega }{1\times 10^{-3}}=30\times 10^{-6}$
$\Rightarrow \omega =\frac{30\times 10^{-6}\times 10^{-3}}{3\times 10^{-12}\times 10^{-7}}$
$=10^{11}rad{s}^{-1}$