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 x direction and has a magnitude of 2.44 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1770 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.82 V/m, (b) in the negative z direction and has a magnitude of 3.82 V/m, and (c) in the positive x direction and has a magnitude of 3.82 V/m?

Explanation / Answer

Force is given by q*VxB (16*10^-20*1770*2.44*10^-3= 6.91 *10^-19 N) and other by qe so, a. F= 16*10^-20*3.82+6.91*10^-19=1.3 *10^-18 b. F= 9.22 *10^-19 c. F= 6.91*10^-19-16*10^-20*3.82= 7.98 *10^-20

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