A proton moves from south to north with a speed of $2 × 10^7 m s^{-1}$. When a magnetic field is applied from west to east, the proton gets an acceleration of $10^{12} m s^{-2}$. The mass of the proton is $1.6 × 10^{-27} kg$. The value of the magnetic field is

$5 × 10^{-4} T$

The correct answer is Option (1) → $5 × 10^{-4} T$

Given:

Velocity of proton: $v = 2 \times 10^{7} \, m/s$ (south → north)

Magnetic field: west → east

Acceleration: $a = 10^{12} \, m/s^2$

Mass of proton: $m = 1.6 \times 10^{-27} \, kg$

Charge of proton: $q = 1.6 \times 10^{-19} \, C$

Magnetic force: $F = q v B$

Also, $F = ma$

So, $q v B = m a$

$B = \frac{ma}{qv}$

Substitute values:

$B = \frac{(1.6 \times 10^{-27})(10^{12})}{(1.6 \times 10^{-19})(2 \times 10^{7})}$

$B = \frac{1.6 \times 10^{-15}}{3.2 \times 10^{-12}}$

$B = 0.5 \times 10^{-3}$

$B = 5 \times 10^{-4} \, T$

Answer: The magnetic field is $5 \times 10^{-4} \, T$.