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

The correct option is A $1.5\times {10}^{3}m/s$

Here, the magnetic force (Bqv) will provide the necessary centripetal force
$\left(\frac{m{v}^2}{r}\right)$
$\therefore Bqv=\frac{m{v}^2}{r}$
$\Rightarrow Bqr=mv$
For electron and proton, the magnetic field B, charge q and radius r, all same.
$\therefore$ mv = constant
i.e. $m_e{v}_e=m_p{v}_p$
$v_P=\left(\frac{m_e}{m_p}\right)v_e=\left(\frac{9\times {10}^{-31}}{1.8\times {10}^{-27}}\right)3\times {10}^6$
$=1.5\times {10}^{3}m/s$