Consider a standard xyz grid If a magnetic field of 0270 T i

Consider a standard x,yz grid. If a magnetic field of 0.270 T is pointing in the positive Y direction, and a 5.87E3 V/m Electric field is pointing in the positive X direction. How fast would a proton need to be going, so that it stays moving on the positive Z direction? 4.60E4 ms 2.17E4 m/s 7.36E4 m/s 3.48E4 m/s

Solution

v = vz k

B = 0.27 j

E = 5.87*10^3 i

the magnetic force Fb = q*(v x B )

Fb = q*vz*B (k x j ) = q*vz*B (-i)

electric force Fe = E*q = Ex*q i

Fnet = Fe + Fb

for constant speed acceleration a = 0

Fnet = m*a

Fe + Fb = 0

q*vz*B = Ex*q

vz = Ex/B = 5.87*10^3/0.27

vz = 2.17E4 m/s <<<<<<-----ANSWER