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

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

Zero

$8.85 \times 10^{- 15} N$ directed out of the plane of the paper

none of the above

answer is A.

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Detailed Solution

Electric filed in between the capacitor plates is given by
$E = \frac{Q}{ε_{0} A}$
Where Q is the charge on capacitor.
$Q = ε_{0} A \times E = 8.85 \times 10^{- 12} \times 25 \times 10^{- 7} \times 400$ $= 8.85 \times 10^{- 15} C$
Magnetic force experienced by +ve plate is,
$F_{m} = Q v B = 8.85 \times 10^{- 15} \times 25 \times 4$
$= 8.85 \times 10^{- 13} N$ in direction out of the plane of paper.