A point charge $Q\left(=3 \times 10^{-12} C \right)$ rotates uniformly in a vertical circle of radius $R(=1 \,mm )$. The axis of the circle is aligned along the magnetic axis of the earth. At what value of the angular speed $\omega$, the eff ective magnetic field at the centre of the circle .............. $rad / s$ will be reduced to zero? (Horizontal component of earth's magnetic field is $30 \,\mu T )$
KVPY 2015, Diffcult
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(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 ficld.
i.e. $\quad B_q=B_H \Rightarrow \frac{\mu_0 i}{2 R}=B_H$
$\Rightarrow \quad \frac{\mu_0\left(\frac{q}{T}\right)}{2 R}=B_H \Rightarrow \frac{\mu_0 q(i)}{2 R \cdot 2 \pi}=B_H$
$\Rightarrow \quad \frac{\mu_0 q(\omega)}{4 \pi R}=B_H$
Here $\frac{\mu_{\bullet}}{4 \pi}=10^{-7} TmA ^{-1}, q=3 \times 10^{-12} \,C$,
$R=1 \,mm =10^{-3} \,m , B_H=30 \times 10^{-6} \,T$
$\Rightarrow \quad \frac{10^{-7} \times 3 \times 10^{-12} \times \omega}{1 \times 10^{-3}}=30 \times 10^{-6}$
$\Rightarrow \quad \omega=\frac{30 \times 10^{-6} \times 10^{-3}}{3 \times 10^{-12} \times 10^{-7}}$
$=10^{11} \,rad s ^{-1}$
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