An electron of mass 0.9 x 10^(-30) kg, under the action of magnetic field moves in a circle of radius 2.0cm at a speed of 3 x 10^ 6 m/s. If the proton of mass 1.8 x 10^(-27) were to move in a circle of same radius in the same magnetic field, find the speed.

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Solution

$F o r e l e c t r o n \frac{m_{e} {v_{e}}^{2}}{r} = e V_{e} B B = \frac{m_{e} v_{e}}{e r} f o r p r o t o n \frac{m_{p} {v_{p}}^{2}}{r} = e V_{p} B V_{p} = \frac{e r B}{m_{p}} = \frac{e r \times}{m_{p}} \frac{m_{e} v_{e}}{e r} = \frac{m_{e} v_{e}}{m_{p}} = \frac{0.9 \times 10^{- 30} \times 3 \times 10^{6}}{1.8 \times 10^{- 27}} v_{p} = 1.5 \times 10^{3} m / s$

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An electron of mass 0.9 x 10^(-30) kg, under the action of magnetic field moves in a circle of radius 2.0cm at a speed of 3 x 10^ 6 m/s. If the proton of mass 1.8 x 10^(-27) were to move in a circle of same radius in the same magnetic field, find the speed.

For electronmeve2r = eVeBB = meveerfor protonmpvp2r = eVpBVp = erBmp =er×mpmeveer = mevemp =0.9×10-30×3×1061.8×10-27vp = 1.5×103m/s