What is the force experienced by an electron moving with velocity of 0.5c in a magnetic field \( \vec{B} \) of 2T? The angle between the velocity vector and \( \vec{B} \) is 90°.
(c is the velocity of light, \( e = 1.6\times10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4.8\times10^{-11}\,\mathrm{N} \)
The magnetic force on a moving charge is \( F = qvB\sin\theta \) (the Lorentz force). The angle is \( 90^\circ \), so \( \sin\theta = 1 \) and the force has its maximum value \( qvB \).
Speed of the electron, taking \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \):
\( v = 0.5c = 1.5\times10^{8}\,\mathrm{m\,s^{-1}} \)
\( F = (1.6\times10^{-19}\,\mathrm{C})\times(1.5\times10^{8}\,\mathrm{m\,s^{-1}})\times(2\,\mathrm{T}) = 4.8\times10^{-11}\,\mathrm{N} \)