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प्रश्न
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.
उत्तर
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`
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