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A point charge $Q=\left(=3 \times 10^{-12} \mathrm{C}\right)$ rotates uniformly in a vertical circle of radius $\mathrm{R}=1 \mathrm{~mm}$. The axis of the circle is aligned along the magnetic axis of the earth. At what value of the angular speed $\omega$, the effective magnetic field at the center of the circle will be reduced to zero? (Horizontal component of Earth's magnetic field is 30 micro Tesla)
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Verified Answer
The correct answer is:
$\mathrm{10}^{11} \mathrm{rad} / \mathrm{s}$
$$
\frac{\mu_{0}}{2 R} \cdot\left(\frac{\omega \cdot q}{2 \pi}\right)=30 \times 10^{-6}
$$
So $\omega=10^{11} \mathrm{rad} / \mathrm{s}$
\frac{\mu_{0}}{2 R} \cdot\left(\frac{\omega \cdot q}{2 \pi}\right)=30 \times 10^{-6}
$$
So $\omega=10^{11} \mathrm{rad} / \mathrm{s}$
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