Discussion Forum

An electron of mass 9.0 x 10^-31 kg under the action of a magnetic field moves in a  circle of radius 2 cm at a speed of 3 x 10⁶ m/s.  If a  proton  of mass 1.8 x 10²⁷ kg was to move in a circle of same radius in the same magnetic field, then the speed will become

Answer:

Explanation:

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

 i.e, $m_{e}v_{e}=m_{p}v_{p}$

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

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