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Question
A proton is projected with a velocity of 3 × 106 m s−1 perpendicular to a uniform magnetic field of 0.6 T. Find the acceleration of the proton.
Solution
Given:
Velocity of the proton, v = 3 × 106 m s−1
Uniform magnetic field, B = 0.6 T
As per the question, the proton is projected perpendicular to a uniform magnetic field.
We know,
F = mpa ....(i)
and
F = evBsinθ ....(ii)
On equating (i) and (ii), we get:
ma = evBsinθ (As θ = 90˚)
`a = (evB)/(m)`
`(1.6xx10^-19xx3xx10^6xx0.6)/(1.67xx10^-27)`
= 1.72 × 1014 m/s2
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