Answer
(a) $0.11$ $T$, vertically upward direction
(b) $0.11$ $T$, vertically downward direction
Work Step by Step
(a) The magnetic fore acting on a particle having charge $q$ and moving with a velocity $(\vec v)$ in a uniform magnetic field $(\vec B)$ is given by $\vec F=q(\vec v\times \vec B)$
As the proton experiences maximum magnetic force, then $\vec v \perp \vec B$.
Therefore, $|\vec F|=qvB$ ................$(1)$
Given,$ |\vec F|=8.0\times 10^{-14}$ $N$, $v=4.5\times 10^{6}$ $m/s$
Charge of a proton $q=1.6\times 10^{-19}$ $C$.
Putting the above values in equation $(1)$, we get
$B=\frac{|\vec F|}{qv}$
or, $B=\frac{8.0\times 10^{-14}}{1.6\times 10^{-19}\times4.5\times 10^{6}}$ $T$
or, $B= 0.11$ $T$
Using right hand rule, we get the direction of the magnetic field, which is in vertically upward direction.
(b) If the proton is replaced by an electron :
(i) the magnitude of the magnetic field remains unchanged because the magnitude of the charge of an electron is identical to a proton
(ii) but the direction of magnetic will be opposite (i.e, vertically downward) because electron is negatively charged which is opposite to a proton.
