A proton moves at 8.00 × 10⁶ m/s along the x-axis. It enters a region in which there is a magnetic field of magnitude 2.50 T, directed at an angle of…
Source: JAMB · 2024
A proton moves at 8.00 × 10⁶ m/s along the x-axis. It enters a region in which there is a magnetic field of magnitude 2.50 T, directed at an angle of 60.0° with the x-axis and lying in the xy-plane. Calculate the proton’s initial acceleration. (charge of proton = 1.60 × 10⁻¹⁹ C, m_p = 1.67 × 10⁻²⁷ kg)
Explanation
Only the part of the field at right angles to the velocity gives a force. Magnetic force F = qvB sinθ.
F = 1.60 × 10⁻¹⁹ × 8.00 × 10⁶ × 2.50 × sin 60° = 3.2 × 10⁻¹² × 0.866 ≈ 2.77 × 10⁻¹² N.
Acceleration = F ÷ m = 2.77 × 10⁻¹² ÷ 1.67 × 10⁻²⁷ ≈ 1.66 × 10¹⁵ m/s².
The 2.77 × 10⁻¹² option is the force in newtons, not the acceleration.
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