In the figure provided, an antiproton (same properties as a proton except that q

Question

In the figure provided, an antiproton (same properties as a proton except that q = -e) is moving in through electric and magnetic fields. If B = 3.31 T, E = 4.78×103 V/m and v = 468 m/s, what are the magnitude and direction of the antiproton’s acceleration at this instant?

Magnitude:

Direction:

Question 2 elds. t -3 31 T E-4,7 x 10, em and V. 4 B m/s, mat are tn° ma nt da and drect on at tne antiprctcr saccalaration :t hls Instant? 0 in thrcu n ortre and magnetic o on e capt hat q in the t ura pravidac an art proton samo ora artios asa Magntude Directian Up Down Rght

Explanation / Answer

Electric force acts down wards

Electric force Fe = -E*q

magnetic force acts upwards

magnetic force Fb = q*v*B

net force = Fe + Fb

Fnet = q*v*B - E*q

from newtons second law

Fnet = m*a

q*v*B - E*q = m*a

acceleration a = e*(vB - E)/m

acceleration a = 1.6*10^-19*(468*3.31 - 4.78*10^3)/(1.67*10^-27)

acceleration a = -3.1*10^11 m/s^2

magnitude = 3.1*10^11 M/S^2

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DIRECTION DOWN

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