$F_2 = |q_2| v_2 B sin theta_2$

If the magnitude of the force is $0.0500:N$, the relative angle $theta_2$ between the direction of the motion of the charge and the magnetic field is given by

$theta_2 = sin^{-1} (dfrac{F_2}{|q_2| v_2 B})$

Substituting the values,

$theta_2 = sin^{-1} (dfrac{0.05}{53 times 10^{-6} times 1.30 times 10^3 times 1.37}) = sin^{-1}(0.5297) = 32^o$

When proton moves in a direction perpendicular to the magnetic field $vec{B}$, the magnitude of the force acting on it is given by

$F = q vB$

where $v$ is the velocity of the proton and $B$ is the magnitude of the magnetic field.

From Newton’s second law of motion,

$m_p a = qvB$

where $m_p = 1.673 times 10^{-27}:kg$ is the mass of the proton and $a$ is its acceleration due to magnetic force.

So the acceleration of the proton can be written as

$a = dfrac{qvB}{m_p}$

If $v= 9.5:m/s$ and $B =1.6:T$ then,

$a = dfrac{1.6 times 10^{-19} times 9.5 times 1.6}{1.673 times 10^{-27}} = dfrac{1.6 times 9.5 times 1.6}{1.673} times 10^8 = 14.54 times 10^8:m.s^{-2}$

If in magnetic field $vec{B}$, an electron moves with velocity $v$ in the direction perpendicular to magnetic field then the magnitude of the force exerted on the electron is given by

$F = |q| v B$

where $B$ is the magnitude of the magnetic field.

If the magnetic field of $0.12 :T$ exerts force of $8.9 times 10^{-15}:N$, then the speed of the electron is given by

$v = dfrac{F}{|q| B} = dfrac{8.9 times 10^{-15}}{1.6 times 10^{-19} times 0.12} = dfrac{8.9}{1.6 times 0.12} times 10^4 = 46.35 times 10^4:m/s$