Question

A charged particle is moving in uniform magnetic field in a circular path. The radius of circular path is $R$. When energy of particle is doubled, then new radius will be :

• A. $ \frac{R}{\sqrt{2}} $
• B. $ 2R $
• C. $ \frac{R}{2} $
• D. $ \sqrt{2}R $

Answer 1

Answer: (D)

Force acting on charged particle moving in a magnetic fields is given by: $F =$ qvB This is equal to the centripetal force acting on the body.
$F =\frac{ mv ^{2}}{ R }$
Hence $R=\frac{m v}{q B}$
Given initial Radius $R =\frac{ mv }{ qB }$
Now energy is doubled.
i.e. $\frac{1}{2} mv _{2}^{2}=2 * \frac{1}{2} mv ^{2}$
i.e. $v _{2}=\sqrt{2} v$
Hence Radius $R _{2}=\frac{\sqrt{2} mv }{ qB }=\sqrt{2} R$.