Question

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

Answer 1

Here, magnetic force will provide the necessary centripetal force.

$Bqv=\frac{m{v}^2}{r}$

$Bqr=mv$

For electron and proton, the magnetic field $B$ and charge $q$ are same, and from question, radius $r$ of the circle is also same. So,

$Bqv=mv=\text{constant}$

$\therefore m_e{v}_e=m_p{v}_p$

$v_p=(\frac{m_e}{m_p})v_e=(\frac{9\times {10}^{-31}}{1.8\times {10}^{-27}})\times 3\times {10}^6$

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