Question

A magnetic field of 2.00×10^−7 T in the positive x-direction exists. Three electrons pass through the origin, one at a time, each with a velocity of 5.00×10^7 m/s. The first electron passes the origin along the x-axis, the second along the y-axis, and the third at an angle of 45° with respect to both the y- and z-axes in the yz-plane. What is the force acting on each electron?

a) 8.99×10^−13 N
b) 1.27×10^−12 N
c) 1.60×10^−12 N
d) 2.26×10^−12 N

Answer 1

The force acting on the first electron moving along the x-axis with the magnetic field is 0 N due to no perpendicular component. The second and third electrons experience a force of 1.60 × 10-12 N as they travel perpendicular to the magnetic field.

Step-by-step explanation:

Force on an Electron in a Magnetic Field

To calculate the force on an electron in a magnetic field, we use the formula F = qvB sin(θ), where F is the force, q is the charge of the electron (-1.60 × 10-19 C), v is the velocity of the electron, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field. For each scenario given in the question:

• For the first electron moving along the x-axis, with the magnetic field also in the x-direction, θ = 0° and sin(θ) = 0, thus F = 0 N (θ = 0° or θ = 180° yields no force).
• For the second electron moving along the y-axis, perpendicular to the magnetic field, θ = 90°. Using the values q = -1.60 × 10-19 C, v = 5.00 × 107 m/s, B = 2.00 × 10-7 T, the force F = qvB would be F = 1.60 × 10-19 C × 5.00 × 107 m/s × 2.00 × 10-7 T = 1.60 × 10-12 N.
• For the third electron moving at 45° in the yz-plane, the plane is perpendicular to the magnetic field, making θ = 90°. The force F would be the same as for the second electron, at 1.60 × 10-12 N.

The correct answer is (c) 1.60 × 10-12 N for the second and third electrons, while the first electron experiences no force.