First slide

Magnetic force

Question

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

Difficult

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

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.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App