MHT CET · Physics · Magnetic Effects of Current
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
- A \(\frac{R}{3}\)
- B \(R\)
- C \(3 R\)
- D \(\sqrt{3} \mathrm{R}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3} \mathrm{R}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Bqv}=\frac{\mathrm{mv}^{2}}{\mathrm{R}} \quad \therefore \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}\)
\(\frac{1}{2} \mathrm{mv}_{2}^{2}=3 \times \frac{1}{2} \mathrm{mv}_{1}^{2}\)
\(\therefore \mathrm{v}_{2}=\sqrt{3} \mathrm{v}_{1}\)
\(\therefore \quad \mathrm{R}_{2}=\frac{\sqrt{3} \mathrm{mv}}{\mathrm{q} \mathrm{B}}=\sqrt{3} \mathrm{R}\)
\(\frac{1}{2} \mathrm{mv}_{2}^{2}=3 \times \frac{1}{2} \mathrm{mv}_{1}^{2}\)
\(\therefore \mathrm{v}_{2}=\sqrt{3} \mathrm{v}_{1}\)
\(\therefore \quad \mathrm{R}_{2}=\frac{\sqrt{3} \mathrm{mv}}{\mathrm{q} \mathrm{B}}=\sqrt{3} \mathrm{R}\)
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