A uniform magnetic field $B = 1.2mT$ is directed vertically upward throughout the volume of a laboratory chamber. A proton ($m = 1.67 \times {10^{ - 27}}kg$) enters the laboratory horizontally from south to north. Calculate the magnitude of centripetal acceleration of the proton if its speed is $3 \times {10^8}m/s$
(A) $3.45 \times {10^{12}}m/{s^2}$
(B) $1.67 \times {10^{12}}m/{s^2}$
(C) $5.25 \times {10^{12}}m/{s^2}$
(D) $2.75 \times {10^{12}}m/{s^2}$

Hint: To find out the magnitude of the centripetal acceleration of the proton, we need to find out the force involved in moving the proton. The force is caused by the uniform magnetic field present. Then using the value of the force we can calculate the value of the centripetal acceleration.

Formula used: In this solution we will be using the following formula,
$\Rightarrow F = q\left( {\vec V \times \vec B} \right)$
Where, $F$ is the force caused by the presence of the magnetic field, $q$ is the charge of the particle, $\vec V$ is the velocity of the particle, $\vec B$ is the strength of the magnetic field.

Complete step by step solution:
The force on a charged particle which is placed in the presence of the magnetic field is given by,
$\Rightarrow F = q\left( {\vec V \times \vec B} \right)$
Here, we see that the expression involves cross product. But, it is given in the question that the proton enters the magnetic field horizontally. Hence, the angle between which the proton enters the magnetic field is zero.
now the value of $\sin 0$ is 0
Thus, the expression becomes,
$\Rightarrow F = qVB$
Here, $F = ma$
Where, $m$ is the mass of the particle, $a$ is the acceleration of the particle in the field.
Then we get,
$\Rightarrow ma = qVB$
On taking the $m$ from the LHS to the RHS we have,
$\Rightarrow a = \dfrac{{qVB}}{m}$
Now, when we substitute the values in the expression to get the acceleration, we get,
$\Rightarrow a = \dfrac{{1.609 \times {{10}^{19}} \times 3 \times {{10}^8} \times 1.2 \times {{10}^{ - 3}}}}{{1.67 \times {{10}^{ - 27}}}}$
On calculating this we get,
$\Rightarrow a = 3.45 \times {10^{12}}m/{s^2}$
Thus, the magnitude of the centripetal acceleration is $3.45 \times {10^{12}}m/{s^2}$.

Note:
When a charged particle enters a uniform magnetic field, its kinetic energy remains constant. The magnetic field always exerts a force perpendicular to the particle's velocity, so the magnitude of the velocity remains constant and so does the kinetic energy.