Question

A point charge Q moving in a uniform magnetic field of 1.42 T experiences a force of 1.08×10⁻¹² N. The velocity of the charge is perpendicular to the magnetic field. If the magnetic field points South and the force points out of the page, then choose the single correct answer. (In this problem we use the points of the compass for directions on the paper with North pointing to the top of the page, and 'into' and 'out of' to indicate directions with respect to the page.)

- Q is positive, moving North.
- Q is negative, moving South.
- Q is positive, moving West.
- Q is negative, moving West.
- Q is positive, moving East.
- Q is negative, moving North.

The speed of the point charge is 4.74×10⁶ m/s. Calculate the magnitude of the charge.

A: 6.80 x 10⁻²⁰
B: 9.05x10⁻²⁰
C: 1.20x10⁻¹⁹
D: 1.60x10⁻¹⁹
E: 2.13x10⁻¹⁹
F: 2.83x10⁻¹⁹
G: 3.76x10⁻¹⁹
H: 5.01x10⁻¹⁹

Answer 1

To calculate the magnitude of the charge, we can use the equation F = qvB sin θ. Substituting the given values, we find that the magnitude of the charge is 1.20 x 10^-19 C.

Step-by-step explanation:

To determine the magnitude of the charge, we can use the equation F = qvB sin θ, where q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. In this case, the force is given as 1.08 x 10-12 N, the velocity is 4.74 x 106 m/s, and the magnetic field strength is 1.42 T. Since the velocity is perpendicular to the magnetic field, the angle θ is 90 degrees, and sin θ is equal to 1. Substituting these values into the equation, we can solve for q:

F = qvB sin θ
1.08 x 10-12 N = q (4.74 x 106 m/s) (1.42 T) (1)
q = 1.20 x 10-19 C

Therefore, the magnitude of the charge is 1.20 x 10-19 C, which corresponds to option C.