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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Aicharya · Accepted Answer

$\mathrm{Bqv}=\frac{\mathrm{mv}^{2}}{\mathrm{R}} \quad 	herefore \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=3 	imes \frac{1}{2} \mathrm{mv}_{1}^{2}$ $	herefore \mathrm{v}_{2}=\sqrt{3} \mathrm{v}_{1}$ $	herefore \quad \mathrm{R}_{2}=\frac{\sqrt{3} \mathrm{mv}}{\mathrm{q} \mathrm{B}}=\sqrt{3} \mathrm{R}$

A charged particle is moving in a uniform magnetic field in a circular path of radius 'R'. When the energy…

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