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 \( 1.5\times10^{-3}\,\mathrm{T} \) exists in the vertically upward direction. What is the acceleration produced? (Use charge on the proton \( 1.6\times10^{-19}\,\mathrm{C} \) and mass of the proton = \( 1.67\times10^{-27}\,\mathrm{kg} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.75\times10^{11}\,\mathrm{m\,s^{-2}} \)
The proton moves horizontally and the field is vertical, so the velocity is perpendicular to the field (\( \theta = 90^\circ \)). The magnetic force is then \( F = qvB \), and by Newton's second law the acceleration is \( F/m \).
\( \displaystyle a = \frac{qvB}{m} \)
\( q = 1.6\times10^{-19}\,\mathrm{C} \)
\( v = 4\times10^{6}\,\mathrm{m\,s^{-1}} \)
\( B = 1.5\times10^{-3}\,\mathrm{T} \)
\( m = 1.67\times10^{-27}\,\mathrm{kg} \)
\( \displaystyle qvB = 1.6\times10^{-19} \times 4\times10^{6} \times 1.5\times10^{-3} = 9.6\times10^{-16}\,\mathrm{N} \)
\( \displaystyle a = \frac{9.6\times10^{-16}}{1.67\times10^{-27}} = 5.75\times10^{11}\,\mathrm{m\,s^{-2}} \)