Question

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

• A. $2\%$
• B. $4\%$
• C. $10\%$
• D. $Noneoftheabove$

Answer 1

The correct option is A $2\%$

we have,

$F=qvB=\frac{m{v}^2}{r}$

$\therefore r=\frac{mv}{qB}$

$KE=\frac{1}{2}m{v}^2$

$\Rightarrow mv=\sqrt{2km}$

$r=\frac{mv}{qB}\Rightarrow r=\frac{\sqrt{2km}}{qB}$

$\Rightarrow r\propto \sqrt{k}$

$\therefore \frac{\Delta r}{r}=\frac{\Delta k}{2k}$

$=2$%