An alpha-particle moves with a speed of \( (5\times10^{5}\,\hat{\imath})\,\mathrm{m\,s^{-1}} \). It enters a region where there is a magnetic field of magnitude 4 T, directed at an angle of 45° to the X-axis and lying in XY plane. The magnitude of magnetic force on alpha particle is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4.5\times10^{-13}\,\mathrm{N} \)
The magnetic (Lorentz) force on a moving charge is \( \vec{F} = q\,\vec{v}\times\vec{B} \), of magnitude \( qvB\sin\theta \), where \( \theta \) is the angle between \( \vec{v} \) and \( \vec{B} \).
The velocity is along the X-axis and the field is at 45° to it, so \( \theta = 45^\circ \). An alpha particle is a helium nucleus, with charge \( q = 2e \).
\( q = 2 \times 1.6\times10^{-19} = 3.2\times10^{-19}\,\mathrm{C} \)
\( \displaystyle F = qvB\sin\theta = 3.2\times10^{-19} \times 5\times10^{5} \times 4 \times \sin 45^\circ \)
\( \displaystyle F = 6.4\times10^{-13} \times 0.707 \approx 4.5\times10^{-13}\,\mathrm{N} \)