A parallel plate capacitor is moving with a velocity of `25 ms^-1` through a uniform magnetic field of 4T as shown in Fig. If the electric field within the capacitor plates is `400NC^-1` and the plate area is `25xx10^-7m^2`, then the magnetic force experienced by the positive charge plate is

Options

A. `8.85xx10^-13N` directed out of the plane of the paper
B. zero
C. `8.85xx10^-15N` directed out of the plane of the paper
D. none of above

Answer

A

Explanation

(a) Electric field in between the capacitor plates is given by
`E=Q/(epsilon_0A)`
where Q is the charge on capacitor.
`Q=epsilon_0AxxE=8.85xx10^-12xx25xx10^-7xx400`
`=8.85xx10^-15C`
Magnetic force experienced by +ve plate is,
`F_m=QvB=8.85xx10^-15xx25xx4`
`=8.85xx10^-13N` in direction out of the plane of
paper.