Question

If the charge is Q = 3.0 x 10⁻⁶C, its speed v = 1,200 m/s, and

the magnetic field strength B = .5T, calculate:
a.) The magnitude of the magnetic force acting on the moving charge.

Answer 1

The magnitude of the magnetic force acting on a moving charge with a charge of Q = 3.0 x 10⁻⁶ C, speed v = 1,200 m/s, and magnetic field strength B = 0.5 T is 1.8 x 10⁻⁳ N, assuming the velocity and magnetic field are perpendicular.

Step-by-step explanation:

To calculate the magnitude of the magnetic force acting on a moving charge, we can use the formula:

F = qvB sin θ

Where:

• F is the magnetic force,
• q is the charge,
• v is the velocity of the charge,
• B is the magnetic field strength,
• θ is the angle between the velocity and the magnetic field.

Given that the charge Q = 3.0 x 10⁻⁶ C, its speed v = 1,200 m/s, and the magnetic field strength B = 0.5 T (assuming the velocity is perpendicular to the magnetic field hence sin(90°) = 1), we can calculate the magnetic force as follows:

F = (3.0 x 10⁻⁶ C)(1,200 m/s)(0.5 T)sin(90°)

F = (3.0 x 10⁻⁶ C)(1,200 m/s)(0.5 T)(1)

F = 1.8 x 10⁻⁳ N

Therefore, the magnitude of the magnetic force acting on the moving charge is 1.8 x 10⁻⁳ N.