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

The charge/mass ratio of the particle is q/m = 2.5 x 10^5 c/kg

Explanation:

Given data:

Diameter of circle "d" = 1 cm

Radius of circle "r" = 0.5 cm  = 0.5 × 10–2 m

Magnetic field "B" = 0.4 T

Electric field "E" = 200 V/m

Solution:

The path will e straight

If qE = quB

=> E = (rqB X B)/m  [Since r = mv/qB]

E = rqB^2/m

Charge/mass ratio  = q/m = E/B^2 r

= 200/(0.4 x 0.4 x 0.5 x 10^-2)

= 2.5 x 10^5 c/kg

Thus the charge/mass ratio of the particle is q/m = 2.5 x 10^5 c/kg

Also learn more

What is charge, mass and charge to mass ratio of an electron?

https://brainly.in/question/6705130

Answer 2

The mass/charge ratio of the particle is $2.5 \times 10^5 C / \mathrm{kg}$.

Explanation:

It is given:  

Circle’s diameter = 1.0 cm  

Circle’s radius,,

Electric field, ,

Magnetic field, B = 0.40 T.

According to the information provided, the particle moves in a circular direction under magnetic field’s action.

But on the particle, when an electric field is applied, it transfers in a straight line.

It can be written as

$q E=\mathrm{qvB}$,  

where the velocity is written as v and the charge is written as q.

As $r=m v q B, q m=v r B=2.5 \times 105 C / \mathrm{kg}$.