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Q. A uniform magnetic field is at right angle to the direction of motion of proton. As a result, the proton describes a circular path of radius 2.5 cm. If the speed of proton is doubled then the radius of the circular path will be :

MGIMS WardhaMGIMS Wardha 2006

Solution:

$ \frac{m{{v}^{2}}}{r}=evB $ $ r=\frac{mv}{eB} $ $ \frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{v}_{2}}}{{{v}_{1}}} $ $ \frac{{{r}_{2}}}{2.5}=\frac{2v}{v} $ $ \Rightarrow $ $ {{r}_{2}}=5\,cm $