Question

A particle of charge 2.0 × 10⁻⁸ C and mass 2.0 × 10⁻¹⁰ g is projected with a speed of 2.0 × 10³ m s⁻¹ in a region with a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.

Answer 1

Given:
Charge of the particle, q = 2.0 × 10⁻⁸ C
Mass of the particle, m = 2.0 × 10⁻¹⁰ g
Projected speed of the particle, v = 2.0 × 10³ m s⁻¹
Uniform magnetic field, B = 0.10 T.
As per the question, the velocity is perpendicular to the field.
So, for the particle to move in a circle,the centrifugal force to the particle will be provided by the magnetic force acting on it.
Using qvB = $\frac{m{v}^2}{r}$, where r is the radius of the circle formed,
$$\begin{gathered} r=\frac{mv}{qB} \\ =\frac{2\times {10}^{-13}\times 2\times {10}^3}{2\times {10}^{-8}\times 0.10} \\ =20\mathrm{cm} \end{gathered}$$
Time period,
$$\begin{gathered} T=\frac{2\mathrm{\pi }m}{qB} \\ =\frac{2\times 3.14\times 2\times {10}^{-13}}{2\times {10}^{-8}\times 0.10} \\ =6.28\times {10}^{-4}\mathrm{s} \end{gathered}$$