Question

An elecgtron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\overrightarrow{B}$. If kinetic energy is doubled and magnetic induction tripled, the radius will become:

• A. $\frac{3R}{2}$
• B. $\sqrt{\frac{3}{2}}R$
• C. $\sqrt{\frac{2}{9}}R$
• D. $\sqrt{\frac{4}{3}}R$

Answer 1

The correct option is C $\sqrt{\frac{2}{9}}R$

Given:$K.E{.}^{{}^{\prime }}=2K.E.,B^{{}^{\prime }}=3B$

Solution: By using the formula of radius of moving charge in circular orbit in presence of magnetic field i.e.$R=\frac{mv}{qB}$

And as we know that,$p=mv=\sqrt{2mK.E.}$

So,$R=\frac{\sqrt{2mK.E.}}{qB}$

Now put values $R^{{}^{\prime }}=\frac{\sqrt{2m\times 2K.E.}}{q\times 3B}$

$R^{{}^{\prime }}=\sqrt{\frac{2}{9}}R$

Hence the correct answer is C