The drawing shows a parallel plate capacitor that is moving with a speed of 43.8 m/s through a 3.73-T magnetic field. The velocity vis perpendicular to the magnetic field. The electric field within the capacitor has a value of 145 N/C, and each plate has an area of9.38 x 10-4 m². What is the magnitude of the magnetic force exerted on the positive plate of the capacitor?Number iSave for LaterUnitsEAttempts: 0 of 3 used Submit Answer

The magnetic force exerted on a charged particle moving in a magnetic field is given by: F = q v B…

Answered: The drawing shows a parallel plate…

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