What is the magnitude of the magnetic force on a charged particle (Q = 5.0 $\mu$C) moving with a speed of 80 km/s in the positive x direction at a point where Bx = 5.0 T, By = -4.0 T, and Bz = 3.0 T?

Question

Quizplus · Accepted Answer

The equation for the magnetic force on a charged particle is:
F = QVB
where Q is the charge of the particle, V is its velocity, and B is the magnetic field.

First, we need to convert the particle's speed from km/s to m/s:
80 km/s = 80,000 m/s

Next, we can plug in the given values for Q, V, and B:
F = (5.0 F1F1F1S1F1F1F10C)(80,000 m/s)(5.0 T)   [for the x-component]
F = (5.0 F1F1F1S1F1F1F10C)(80,000 m/s)(F1F1F1S1F1F1F104.0 T)   [for the y-component]
F = (5.0 F1F1F1S1F1F1F10C)(80,000 m/s)(3.0 T)   [for the z-component]

Simplifying each of these equations, we get:
Fx = 2.0 N
Fy = -4.0 N
Fz = 1.2 N

The magnitude of the force is given by:
|F| = sqrt(Fx^2 + Fy^2 + Fz^2)

Plugging in the values we just calculated, we get:
|F| = sqrt((2.0 N)^2 + (-4.0 N)^2 + (1.2 N)^2)
|F| = sqrt(4.0 N^2 + 16.0 N^2 + 1.44 N^2)
|F| = sqrt(21.44 N^2)
|F| = 4.62 N

Rounding to the nearest tenth, we get:
|F| = 4.6 N

Therefore, the answer is D) 2.0 N.

What Is the Magnitude of the Magnetic Force on a Charged