Question

A charged particle moves in a circular orbit of radius $r$ in a uniform magnetic field $B_0$. If the kinetic energy is doubled and the magnetic field is tripled, then the radius of the circular orbit will be:

• 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

$\sqrt{\frac{2}{9}}R$

Step 1: Given

The radius of orbit of the particle is $R$

Magnetic Field is $\hspace{0.17em}{B}_0$

Kinetic Energy is $K$

Step 2: Formula Used

The radius of orbit of a particle is given by the formula $R=\frac{mv}{qB}$, where $m$ is mass of the particle, $v$ is velocity of the particle, $q$ is charge of the particle and $B$ is the magnetic field.

And the kinetic energy of the particle is given by the formula $K=\frac{1}{2}m{v}^2$.

By both these formulas it can be concluded that $R=\frac{\sqrt{2mK}}{qB}$.

Step 3: Solution

So, the radius of the particle for magnetic field $B_0$ is

$R=\frac{\sqrt{2mK}}{q{B}_0}$

Replace the kinetic energy $K$ by $2K$ and the magnetic field $B_0$ by $3B_0$

$\begin{array}{rcl}R& =& \frac{\sqrt{2mK}}{q{B}_0}\\ & =& \frac{\sqrt{2m\left(2K\right)}}{q\left(3B_0\right)}\\ & =& \frac{\sqrt{2}}{3}\frac{\sqrt{2mK}}{q{B}_0}\\ & =& \sqrt{\frac{2}{9}}R\end{array}$

So, the radius will be $\sqrt{\frac{2}{9}}R$.

The correct option is (C).