The drawing shows a parallel plate capacitor that is moving with a speed of 40 m

Question

The drawing shows a parallel plate capacitor that is moving with a speed of 40 m/s through a 4.4-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 240 N/C, and each plate has an area of 9.6

The drawing shows a parallel plate capacitor that is moving with a speed of 40 m/s through a 4.4-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 240 N/C, and each plate has an area of 9.6 times 10-4 m2. What is the magnitude of the magnetic force exerted on the positive plate of the capacitor?

Explanation / Answer

E=q/eoA

=>q=eo*E*A =(8.85*10^-12)*240*(9.6*10^-4)

q=2.04*10^-12 C

magnetic force exerted

F=qVB =2.04*10^-12*40*4.4

F=3.59*10^-10 N =3.6*10^-10 N (approx)

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