Question

An electron that has a velocity with x component 2.0 × 106 m/s and y component 3.0 x 106 m/s moves through a uniform magnetic field with x component 0.024 T and y component -0.12 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

Answer 1

The magnetic force on a charged particle moving through a magnetic field is given by the following formula:

F = q * v * B * sin(theta)

where
- F is the magnetic force
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnetic field
- theta is the angle between the velocity and the magnetic field

(a) For an electron, q = -1.6 x 10^-19 C, v = (2.0 x 10^6) i + (3.0 x 10^6) j m/s, B = (0.024) i - (0.12) j T, and theta = 90 degrees (since the velocity is perpendicular to the magnetic field).

Plugging in the values, we get:

F = (-1.6 x 10^-19) * ((2.0 x 10^6) i + (3.0 x 10^6) j) * ((0.024) i - (0.12) j) * sin(90 degrees)

F = 4.8 x 10^-13 N

Therefore, the magnitude of the magnetic force on the electron is 4.8 x 10^-13 N.

(b) For a proton, q = 1.6 x 10^-19 C (since the proton has a positive charge), and all other values are the same as for the electron.

Plugging in the values, we get:

F = (1.6 x 10^-19) * ((2.0 x 10^6) i + (3.0 x 10^6) j) * ((0.024) i - (0.12) j) * sin(90 degrees)

F = -1.2 x 10^-14 N

Therefore, the magnitude of the magnetic force on the proton is 1.2 x 10^-14 N.