Answer
The magnitude of the magnetic field is $0.117$ $T$ and the direction is parallel or anti-parallel to the velocity of the charged particle.
Work Step by Step
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
$\vec F=q(\vec v\times \vec B)$
or, $|\vec F|=qvB\sin\theta$, where $\theta$ is angle between $\vec v$ and $\vec B$
According to the problem, the particle experiences no magnetic force, although there is a magnetic field present. It is possible only when the direction of the magnetic field is parallel or anti-parallel to the velocity of the charged particle.
It is also given that the maximum possible magnetic force that the charge could experience has a magnitude of 0.48 N. it is possible when the direction of the magnetic field is perpendicular to the velocity of the charged particle.
Thus, $|\vec F|=qvB$
or, $B=\frac{|\vec F|}{qv}$
or, $B=\frac{0.48}{(8.2\times 10^{-6})\times (5.0\times 10^{5})}$ $T$
or, $B=0.117$ $T$
Therefore, the magnitude of the magnetic field is $0.117$ $T$ and the direction is parallel or anti-parallel to the velocity of the charged particle.
