Question

A charged particle moving in a uniform magnetic field when looses $3\%$ of its kinetic energy then the radius of curvature of its circular path.

• A. Increases by $15\%$
• B. Decreases by $4.5\%$
• C. Decreased by $1.5\%$
• D. Increases by $3\%$

Answer 1

Answer: (C)

Decreased by $1.5\%$

The radius of charged particle moving in uniform magnetic field is
$r=\frac{mv}{qB}$
$\Rightarrow r^2=\frac{m^2{v}^2}{q^2{B}^2}=\frac{2m\left(\frac{1}{2}m{v}^2\right)}{q^2{B}^2}=\frac{2mk}{q^2{B}^2}$
$\therefore r=\frac{\sqrt{2mk}}{qB}$
$r\propto \sqrt{K}$
$\frac{\Delta r}{r}=\frac{1}{2}\frac{\Delta K}{K}$
$\therefore \frac{\Delta r}{r}\times 100=\frac{1}{2}\left(\frac{\Delta K}{K}\times 100\right)$
$=\frac{1}{2}\times 3=1.5\%$.