Question

A charged particle moving in a uniform magnetic field when losses $4\%$ of its kinetic energy, the radius of curvature of its circular path:

• A. Decreases by $2\%$
• B. Increases by $2\%$
• C. Increases by $4\%$
• D. Decreases by $4\%$

Answer 1

The correct option is A Decreases by $2\%$

We know the radius of a circular path of charge $q$ moving with kinetic energy $K$ in a uniform magnetic field is given by,

$r=\frac{\sqrt{2mK}}{qB}$

As $m,q$ and $B$ are the same for the same charge.

$\therefore r\propto \sqrt{K}$

$r=C{K}^{\frac{1}{2}}$ (where, $C$ is a constant)

Taking $\log$ on both sides, we get

$\log r=\log C+\frac{1}{2}\log K$

Differentiating both sides;

$\frac{1}{r}\times dr=0+\frac{1}{2}\left(\frac{1}{K}\right)dK$

$\frac{dr}{r}\times 100=\frac{1}{2}\times \frac{dK}{K}\times 100$

As given, $\frac{dK}{K}\times 100=4\%$

$\therefore \frac{dr}{r}\times 100=\frac{1}{2}\times 4=2\%$

Thus, radius will be decreasing by $2\%$.

Hence, option $\left(a\right)$ is the correct answer.