A proton is projected with a speed of $4 × 10^6 m s^{-1}$ horizontally from east to west. A uniform magnetic field of strength $1.5 × 10^{-3} T$ exists in the vertically upward direction. What is the acceleration produced? (Use charge on the proton $1.6 × 10^{-19} C$ and mass of the proton = $1.67 × 10^{-27} kg$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $5.75 × 10^{11} ms^{-2}$
Given:
Speed of proton, $v = 4 \times 10^{6}\,m/s$
Magnetic field, $B = 1.5 \times 10^{-3}\,T$
Charge on proton, $q = 1.6 \times 10^{-19}\,C$
Mass of proton, $m = 1.67 \times 10^{-27}\,kg$
Magnetic force on a moving charge:
$F = qvB \sin\theta$
Here, velocity is horizontal (east–west) and field is vertical (upward), so $\theta = 90^\circ$, $\sin\theta = 1$.
$F = 1.6 \times 10^{-19} \times 4 \times 10^{6} \times 1.5 \times 10^{-3}$
$F = 1.6 \times 4 \times 1.5 \times 10^{-16}$
$F = 9.6 \times 10^{-16}\,N$
Acceleration:
$a = \frac{F}{m} = \frac{9.6 \times 10^{-16}}{1.67 \times 10^{-27}}$
$a = 5.75 \times 10^{11}\,m/s^2$
Final Answer: $a = 5.8 \times 10^{11}\,m/s^2$