Answer
$1.1\times 10^{-4}$ $T$
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, $|\vec F|=qvB \sin\theta$
or, $B=\frac{|\vec F|}{qv\sin\theta}$
or$B=\frac{5.4\times 10^{-3}}{(8.3\times 10^{-6})\times (7.4 \times 10^{6})\times (\sin52^{\circ})}$ $T$
or, $B=1.1\times 10^{-4}$ $T$
Therefore, the magnitude of the magnetic field is $1.1\times 10^{-4}$ $T$
