First slide

Force on a charged particle moving in a magnetic field

Question

An electron of mass 0.90 x 10^-30 kg under the action of magnetic field moves in a circle of radius 2.0 cm at a speed of 3.0 x 10⁶ m/sec. If a proton of mass 1.8 x 10^-27 kg were to move in a circle of the same radius in the same magnetic field, then its speed will be

Moderate

A

3.0 x10⁶m,/sec
B

1 .5 x 10³ m/sec
C

6x10⁴ m/sec
D

cannot be estimated from the above data

Solution

We know that $r = \frac{mv}{qB}$
$∴ \frac{r_{p}}{r_{e}} = \frac{m_{p}}{m_{e}} \times \frac{v_{p}}{v_{e}} (∵ q_{p} = q_{e})$
Given that r_p = r_e
$\begin{matrix} ∴ v_{p} = v_{e} \times (\frac{m_{e}}{m_{p}}) = \frac{(3 \cdot 0 \times 10^{6}) (0 \cdot 90 \times 10^{- 30})}{1 \cdot 8 \times 10^{- 27}} \\ = 1 \cdot 5 \times 10^{3} m / s \end{matrix}$

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