Chapter 20, Problem 21P

To determine

To show: Kinetic energy of the particle is proportional to r² .

Explanation of Solution

Given:

Mass of particle is m

In a magnetic field B ,charge q moves in a circular path whose radius of curvature is r.

Formula used:

The kinetic energy is given as,

  $K.E.=\frac{1}{2}m{v}^2$

Here, m is mass and v is velocity

Calculation:

Magnetic force supplies the centripetal force,

  $F=\frac{m{v}^2}{r}$

Here, r is the radius of curvature

Also, force on a charged particle in the magnetic field is qvB .

So,

  $qvB=\frac{m{v}^2}{r}$

  $r=\frac{mv}{qB}$

  $mv=rqB$

  $v=\frac{rqB}{m}$  (1)

Substitute the value of v from equation (1) in $K.E.=\frac{1}{2}m{v}^2$

  $\begin{array}{c}K.E.=\frac{1}{2}m\left(\frac{rqB}{m}\right)\\ =\frac{1}{2}m\frac{r^2{q}^2{B}^2}{m^2}\\ =\frac{r^2{q}^2{B}^2}{m}\end{array}$

Conclusion:

Thus, the kinetic energy is proportional to square of the radius of curvature.