A proton is projected with a speed of \( 4\times10^{6}\,\mathrm{m\,s^{-1}} \) horizontally from east to west. A uniform magnetic field of strength \( 3.0\times10^{-3}\,\mathrm{T} \) exists in the vertically upward direction. The acceleration of the proton is
(Given: mass of proton \( 1.67\times10^{-27}\,\mathrm{kg} \); charge on proton \( 1.6\times10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 11.5\times10^{11}\,\mathrm{m\,s^{-2}} \)
A moving charge in a magnetic field feels the magnetic Lorentz force \( F = qvB\sin\theta \). The velocity is horizontal and the field is vertical, so \( \theta = 90^\circ \) and \( \sin\theta = 1 \).
\( \displaystyle F = qvB = 1.6\times10^{-19} \times 4\times10^{6} \times 3.0\times10^{-3} = 1.92\times10^{-15}\,\mathrm{N} \)
By Newton's second law,
\( \displaystyle a = \frac{F}{m} = \frac{1.92\times10^{-15}\,\mathrm{N}}{1.67\times10^{-27}\,\mathrm{kg}} \approx 1.15\times10^{12}\,\mathrm{m\,s^{-2}} = 11.5\times10^{11}\,\mathrm{m\,s^{-2}} \)
Gravity (\( 9.8\,\mathrm{m\,s^{-2}} \)) is negligible beside this.