A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative Times direction and has a magnitude of 3.24 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1640 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is in the positive z direction and has a magnitude of 3.69 V/m, in the negative z direction and has a magnitude of 3.69 V/m, and in the positive Times direction and has a magnitude of 3.69 V/m?

Explanation / Answer

As we know that F= m(V×B)+q* E

A)F=1.673*10^-27*1640*3.24*10^-3z hat + 1.6*10^-19*3.69 z hat = 8889.6528*10^-30+5.904*10^-19 z hat

B)F= (8889.6528*10^-30)-5.904*10^-19 z hat

C)F=8889.6528*10^-30 z hat + 5.904*10^-19 x hat

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