Answer
$1.30\times 10^{-10}$ $N$, directed out of the plane of the paper.
Work Step by Step
Consider a parallel-plate capacitor with plate area $A$ and separation $d$ between the plates. Suppose, a charge $+Q$ is given to the positive plate and $-Q$ to the negative plate of the capacitor. The capacitance $C$ of such a parallel plate capacitor is expressed by the formula: $C=\frac{\epsilon_{0}A}{d}$ ....................$(1)$
The potential difference between the plates is
$V=\frac{Q}{C}$ ....................$(2)$
If $E$ be the magnitude of electric field between the plates, then the potential difference between the plates can be written as: $V=Ed$ ....................$(3)$
Putting equation $(3)$ in equation $(2)$, we get
$Ed=\frac{Q}{C}$
or, $C=\frac{Q}{Ed}$ ....................$(4)$
Putting equation $(4)$ in equation $(1)$, we get
$\frac{Q}{Ed}=\frac{\epsilon_{0}A}{d}$
or, $Q=\epsilon_{0}AE$
or, $Q=1.128\times 10^{-12}$ $C$
The velocity $\vec v$ of the positive plate is perpendicular to the magnetic field $\vec B$
Therefore, the magnitude of the the magnetic force exerted on the positive plate of the capacitor is
$|\vec F|=QvB$
$|\vec F|=(1.128\times 10^{-12})\times 32 \times 3.6$ $N$
$|\vec F|=1.30\times 10^{-10}$ $N$
Thus the magnitude of the magnetic force on the positive plate is $1.30\times 10^{-10}$ $N$ and using right hand rule, we get the direction of magnetic force which is directed out of the plane of the paper.
