Question

An electron of mass $9.0 \times 10^{-31} kg$ under the action of a magnetic field moves in a circle of radius $2\, cm$ at a speed of $3 \times 10^6 \, m/s$ . If a proton of mass $1.8 \times 10^{27} \, kg$ was to move in a circle of same radius in the same magnetic field, then its speed will become

• A. $1.5 \times 10^3 \, m/s$
• B. $3 \times 10^6 \, m/s$
• C. $6 \times 10^4 \, m/s$
• D. $2 \times 10^6 \, m/s$

Answer 1

Answer: (A)

$1.5 \times 10^3 \, m/s$

Answer 2

Here, the magnetic force (Bqv) will provide the necessary centripetal force
$\left(\frac{mv^{2}}{r}\right)$
$\therefore \, \, Bqv = \frac{mv^{2}}{r}$
$\Rightarrow Bqr = mv$
For electron and proton, the magnetic field B, charge q and radius r, all same.
$\therefore \, \,$ mv = constant
i.e., $m_{e}v_{e} = m_{p}v_{p}$
$v_{p} = \left(\frac{m_{e}}{m_{p}}\right) v_{e} = \left(\frac{9 \times10^{-31}}{1.8\times 10^{-27}}\right) 3\times 10^{6}$
$= 1.5 \times 10^{3} ms$