Question

A charged particle moving in a uniform magnetic field and losses $4\%$ of its kinetic energy. The radius of curvature of its path changes by

• A. 2 %
• B. 4 %
• C. 10 %
• D. 12 %

Answer 1

Answer: (A)

As we know $F=qvB=\frac{m{v}^2}{r}$
$\therefore r=\frac{mv}{Bq}$
And $KE=k=\frac{1}{2}m{v}^2$
$\therefore mv=\sqrt{2km}$
$\therefore r=\frac{mv}{qB}==\sqrt{\frac{2km}{qB}}$
$\Rightarrow r\propto \sqrt{k}$
or $r=c^{1\backslash 2}$ (cis a constant)
$\frac{dr}{dr}=c\frac{d{k}^{12}}{dr}$
or $\frac{c\Delta k}{\Delta r}=2\sqrt{k}$
or $\frac{\Delta r}{r}=\frac{c\Delta k}{2\sqrt{kc}\sqrt{k}}=\frac{\Delta k}{2k}$
There fore percentage changes in radius of path,
$\frac{\Delta r}{r}$ $\times 100=\frac{\Delta k}{2k}\times 100=2\%$