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 $3.0 × 10^{-3} T$ exists in the vertically upward direction. The acceleration of the proton is

(Given: mass of proton $1.67 × 10^{-27} kg$; charge on proton $1.6 × 10^{-19} C$)

$11.5 × 10^{11}\, m s^{-2}$

The correct answer is Option (2) → $11.5 × 10^{11}\, m s^{-2}$

$a = \frac{qvB}{m}$

$q = 1.6 \times 10^{-19}, \quad v = 4 \times 10^{6}, \quad B = 3 \times 10^{-3}$

$a = \frac{1.6 \times 10^{-19} \cdot 4 \times 10^{6} \cdot 3 \times 10^{-3}}{1.67 \times 10^{-27}}$

$a = \frac{1.92 \times 10^{-15}}{1.67 \times 10^{-27}}$

$a \approx 1.15 \times 10^{12} \ \text{m/s}^2$

$\text{Direction: } \vec{v} \times \vec{B}$

$\text{West} \times \text{Up} \Rightarrow \text{South}$

The acceleration is $1.15 \times 10^{12} \ \text{m/s}^2$ towards south.