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 = $1.67 × 10^{-27} kg$)

$5.75 × 10^{11} ms^{-2}$

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: $a = 5.8 \times 10^{11}\,m/s^2$