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.

Answer 1

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 = m_pa ....(i)
and
F = evBsinθ ....(ii)
On equating (i) and (ii), we get:
ma = evBsinθ (As θ = 90˚)
$$\begin{gathered} a=\frac{evB}{m} \\ =\frac{1.6\times {10}^{-19}\times 3\times {10}^6\times 0.6}{1.67\times {10}^{-27}} \end{gathered}$$
= 1.72 × 10¹⁴ m/s²