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A charged particle is moving in a uniform magnetic field in a circular path of radius 'R'. When the energy of the particle becomes three times the original, the new radius will be
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The correct answer is:
$\sqrt{3} \mathrm{R}$
$\mathrm{Bqv}=\frac{\mathrm{mv}^{2}}{\mathrm{R}} \quad \therefore \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$
$\frac{1}{2} \mathrm{mv}_{2}^{2}=3 \times \frac{1}{2} \mathrm{mv}_{1}^{2}$
$\therefore \mathrm{v}_{2}=\sqrt{3} \mathrm{v}_{1}$
$\therefore \quad \mathrm{R}_{2}=\frac{\sqrt{3} \mathrm{mv}}{\mathrm{q} \mathrm{B}}=\sqrt{3} \mathrm{R}$
$\frac{1}{2} \mathrm{mv}_{2}^{2}=3 \times \frac{1}{2} \mathrm{mv}_{1}^{2}$
$\therefore \mathrm{v}_{2}=\sqrt{3} \mathrm{v}_{1}$
$\therefore \quad \mathrm{R}_{2}=\frac{\sqrt{3} \mathrm{mv}}{\mathrm{q} \mathrm{B}}=\sqrt{3} \mathrm{R}$
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