Question

A charged particle is moving in a uniform magnetic field in a circular path of radius R. When energy of the particle is doubled, then the new radius will be :

• A. $ R $
• B. $ R\sqrt{2} $
• C. $ R/\sqrt{2} $
• D. $ 2R $

Answer 1

Answer: (B)

Energy of charged particle revolving in a magnetic field, $ E=\frac{1}{2}m{{\upsilon }^{2}} $ But $ \frac{m{{\upsilon }^{2}}}{R}=q\upsilon B $ $ \Rightarrow $ $ \upsilon =\frac{qBR}{m} $ $ \therefore $ $ E=\frac{1}{2}m\frac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{{{m}^{2}}} $ $ =\frac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2m} $ $ \Rightarrow $ $ E\propto {{R}^{2}} $ $ \Rightarrow $ $ R\propto \sqrt{E} $ If R' is new radius $ \frac{R'}{R}=\sqrt{\frac{E'}{E}} $ $ \Rightarrow $ $ \frac{R'}{R}=\sqrt{2} $ $ \Rightarrow $ $ R'=\sqrt{2}R $