Question

At one instant an electron (charge = −1.6 ᴇ −19 C) is moving in the xy plane, the components of its velocity being 5 ᴇ 5 m/s in the x-direction and 3 ᴇ 5 m/s in the y-direction. A magnetic field of 0.8T is in the positive x direction. At that instant the magnitude of the magnetic force on the electron is:

A) 0
B) 2.6 × 10−14 N
C) 3.8 × 10−14 N
D) 6.4 × 10−14 N
E) 1.0 × 10−13 N

Answer 1

The magnitude of the magnetic force on the electron is 3.8 ᴇ -14 N, calculated using the formula F = qvBsin(θ) for the y-component of the velocity since it is perpendicular to the magnetic field.

Step-by-step explanation:

To calculate the magnetic force on an electron moving in the xy-plane under the influence of a magnetic field in the positive x-direction, we can use the formula F = qvBsin(θ), where q is the charge of the electron, 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. Since the magnetic field is in the x-direction, only the velocity component in the y-direction will contribute to the force. Therefore, we only consider the velocity component of 3 ᴇ 5 m/s in the y-direction. The angle θ between the velocity in the y-direction and the magnetic field in the x-direction is 90 degrees (or π/2 radians).

The charge of the electron is -1.6 ᴇ -19 C, and the magnetic field is given as 0.8 T. Plugging these values into the equation, we get F = (1.6 ᴇ -19 C)(3 ᴇ 5 m/s)(0.8 T)sin(90°), which simplifies to F = (1.6 ᴇ -19 C)(3 ᴇ 5 m/s)(0.8 T)(1) since sin(90°) = 1. Calculating this product gives us the magnetic force F = 3.84 ᴇ -14 N, which would be rounded to 3.8 ᴇ -14 N to two significant figures.