An electron of mass 9 .0 × 10 - 31 kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of 3 × 10 6 m / s . If a proton of mass 1 .8 × 10 - 27 kg has to move in a circle

Question

An electron of mass 9 .0 × 10 - 31 kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of 3 × 10 6 m / s . If a proton of mass 1 .8 × 10 - 27 kg has to move in a circle of same radius and in the same magnetic field, then its speed must be

TestHub · Accepted Answer

Magnetic force will provide the necessary centripetal force.

$B q v = \frac{m v^{2}}{r}$

$B q r = m v$

For electron and proton, the magnetic field $B$ , charge $q$ and radius $r$ are same.

So, $m v$ = constant

i.e. $m_{e} v_{e} = m_{p} v_{p}$

$v_{p} = (\frac{m_{e}}{m_{p}}) v_{e} = (\frac{9 \times 10^{- 31}}{1.8 \times 10^{- 27}}) \times 3 \times 10^{6}$

$v_{p} = 1.5 \times 10^{3} m / s$

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