Question

An antiproton (same properties as a proton except that q=−e) is moving in the combined electric and magnetic fields of the figure.(Figure 1) Assume that B = 2.3 T and E = 800 V/m . What is the magnitude of the antiproton's acceleration at this instant? What is the direction of the antiproton's acceleration at this instant?

Answer 1

Magnetic force, Fb = q ( v x B)

= (-1.6 x 10^-19) ( 500 x 2.3)

= 1.84 x 10^-16 N

and direction of magetic force will be in the direction of field.

Electric force, Fe = q E = - 1.6 x 10^-19 x 800 = 1.28 x 10^-16 N

q is negative, hence Fe will be in opposite directin of field.

fnet = Fb - Fe

= 5.6 x 10^-17 N

a = fnet / m = (5.6 x 10^-17 ) / (1.67 x 10^-27) = 3.353 x 10^10 m/s^2

in the direction of electric field.