Question

A parallel plate capacitor is moving with a velocity of $25m$ $s^{-1}$ through a uniform magnetic field of $4.0T$ as shown in figure. If the electric field within the capacitor plates is $400N{C}^{-1}$ and plate area is $25\times {10}^{-7}{m}^2$, then the magnetic force experienced by the positive charge plate is

• A. $8.85\times {10}^{-13}N$ directed out of the plane of the paper
• B. zero
• C. $8.85\times {10}^{-15}N$ directed out of the plane of the paper
• D. none of the above

Answer 1

Answer: (A)

$8.85\times {10}^{-13}N$ directed out of the plane of the paper

Electric field in between the capacitor plates is given by

$E=\frac{Q}{{\epsilon }_{0}A}$

where $Q$ is the charge on capacitor

$Q={\epsilon }_{0}A\times E=8.85\times {10}^{-12}\times 25\times {10}^{-7}\times 400$

$=8.85\times {10}^{-15}C$

Magnetic force experienced by +ve plate is,

$F_m=QvB=8.85\times {10}^{-15}\times 25\times 4$

$=8.85\times {10}^{-13}N$ in directed out of the plane of paper.