Question

A parallel plate capacitor is moving with a velocity of $25m{s}^{-1}$ through a uniform magnetic field of 1.5 T as shown in figure. If the electric field within the capacitor plates is $175N{C}^{-1}$ and plate area is $25\times {10}^{-7}{m}^2,$ then the magnetic force experienced by positive charge plate is

• A. $1.45\times {10}^{-13}N$
• B. zero
• C. $8.67\times {10}^{-15}N$
• D. $3.87\times {10}^{-15}N$

Answer 1

Answer: (A)

Electric field in between the capacitor plates is given by $E=\frac{q}{{\epsilon }_{0}A}$ or $q={\epsilon }_{0}A\times E$ $=8.85\times {10}^{-12}\times 25\times {10}^{-7}\times 175$ $=3.87\times {10}^{-15}C$ Magnetic force experienced by +ye plate is, $F_m=qvB=1.45\times 10{}^{-13}N$ in a direction out of plane of paper.