Question

A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V m⁻¹ makes the path straight. Find the charge/mass ratio of the particle.

Answer 1

Given:
Diameter of the circle = 1.0 cm
Thus, radius of circle, r = = 0.5 × 10⁻² m,
Magnetic field, B = 0.40 T
Electric field, E = 200 V m⁻¹.
As per the question, the particle is moving in a circle under the action of a magnetic field. But when an electric field is applied on the particle, it moves in a straight line.
So, we can write:
F_e = F_m
qE = qvB, where q is the charge and v is the velocity of the particle.
$$\begin{gathered} \Rightarrow v=\frac{E}{B}=\frac{200}{0.4}=500\mathrm{m}/\mathrm{s}. \\ \mathrm{As}r=\frac{mv}{qB}, \\ \frac{q}{m}=\frac{v}{rB} \\ =\frac{500}{0.5\times {10}^{-2}\times 0.4} \\ =2.5\times {10}^5\mathrm{C}/\mathrm{kg} \end{gathered}$$