An electron moving with a speed of 4.6×105 m/s in the positive x direction experiences zero magnetic force. When it moves in the positive y direction, it experiences a force of 3.6×10−13 N that points in the positive z direction. Find the components of the magnetic field (Bx, By, Bz)

B = (4.89, 0, 0)

Explanation:

Given Vy = 4.6×10⁵m/s, q = 1.6×10-¹⁹C

and Fz = 3.6×10-¹³N

The magnetic field vector acts along the positive x direction since there was no magnetic force on the electron when it moved in the positive x direction. This is because the equation relating F, q, v and B tells us that

F = qvBSinθ and θ is the angle between B and v that is the direction of the motion of the charged particle. So if θ = 0° that is V and B are in tge same durection, then F = 0 which was the case when the electron moved in the positive x direction. When it moved in the positive y direction it experienced a force in the positiv z direction which validates the assumption that the B vector acts along the positive x durection.

θ = 90°

Therefore,

Bx = F/ qvsinθ = (3.6 × 10-¹³)/ (1.6 × 10-¹⁹ × 4.6×10⁵)×sin90°

Bx = 4.89T

B = (4.89, 0, 0)T.

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