A charged particle is moving horizontally westward with a velocity of 3.5 $\times$ 10⁶ m/s in a region where there is a magnetic field of magnitude 5.6 $\times$ 10^-5 T directed vertically downward.The particle experiences a force of 7.8 $\times$ 10^-16 N northward.What is the charge on the particle? A)+4.0 $\times$ 10^-18 C B)-4.0 $\times$ 10^-18 C C)+4.9 $\times$ 10^-5 C D)-1.2 $\times$ 10^-14 C E)+1.4 $\times$ 10^-11 C

Question

Quizplus · Accepted Answer

The magnitude of the magnetic force acting on a charged particle moving in a magnetic field is given by:

F = qvB sin(theta)

Where q is the charge of the particle, v is the velocity, B is the magnetic field, and theta is the angle between the velocity and magnetic field.

In this problem, the particle is moving horizontally westward, so the angle between its velocity and the downward-directed magnetic field is 90 degrees. Therefore, sin(theta) = 1.

We are given the velocity, magnetic field, and the magnitude of the magnetic force (7.8×10^-16 N) acting on the particle. Therefore, we can solve for the charge (q) using the above equation:

q = F / (vB)

Plugging in the values, we get:

q = (7.8×10^-16 N) / (3.5×10^6 m/s × 5.6×10^-15 T)
q = -4.0×10^-18 C

The negative sign indicates that the charge on the particle is negative, so the best answer choice is B.

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