The drawing shows a parallel plate capacitor that is moving with a speed of 42 m

Question

The drawing shows a parallel plate capacitor that is moving with a speed of 42 m/s through a 4.8-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 210 N/C, and each plate has an area of7.7

The drawing shows a parallel plate capacitor that is moving with a speed of 42 m/s through a 4.8-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 210 N/C, and each plate has an area of7.7 times 10-4 m2. What is the magnitude of the magnetic force exerted on the positive plate of the capacitor

Explanation / Answer

Q=eoEA =(8.85*10^-12)*210*7.7*10^-4

Q=1.43*10^-12 C

F=QVB =1.43*10^-12*42*4.8

F=2.885*10^-10 N

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