Question

A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 10⁶ m s⁻¹. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.

Answer 1

Given:
Applied potential difference, V = 12 kV = 12 × 10³ V
Speed of a charged particle, v =1.0 × 10⁶ m s⁻¹
Magnetic field strength, B = 0.2 T
As per the question, a charged particle is injected perpendicularly into the magnetic field.
We know:
`qV = 1/2 mv^2`

⇒ `m/q = 1/2 mv^2`

= `(2xx12xx10^3)/(1xx10^6)^2`

= 24 ×10^-9 

and r = `(mv)/(qB)`

⇒ r = `(24xx10^-9xx10^6)/(0.2)`

⇒ r = `12xx10^-2m = 12cm`