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)

2%

Answer 2

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