Question

The magnetic field in a region of space is given by B=0.500 T. The velocity vector of an electron moving at 7:49 x 100 m/s makes an angle of 45.0° with this field. What are the magnitudes of the following?

The magnetic force on the electron___________
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N

Answer 1

The magnitude of the magnetic force on an electron moving at 7.49 x 10^6 m/s and making a 45.0° angle with a magnetic field of 0.500 T is approximately 6.38 x 10^-13 N.

Step-by-step explanation:

The magnitude of the magnetic force on an electron can be calculated using the formula: F = qvBsinθ, where q is the charge of the electron, v is the magnitude of its velocity vector, B is the magnitude of the magnetic field, and θ is the angle between the velocity vector and the magnetic field vector.

In this case, the charge of the electron is known to be -1.6 x 10^-19 C, the magnitude of its velocity vector is 7.49 x 10^6 m/s, and the magnetic field is 0.500 T. The angle between the velocity vector and the magnetic field vector is 45.0°.

Plugging in these values into the formula, we get: F = (-1.6 x 10^-19 C)(7.49 x 10^6 m/s)(0.500 T)sin(45°).

Calculating this expression, the resulting magnitude of the magnetic force on the electron is approximately 6.38 x 10^-13 N.