Question

A particle that has an 6.7⋅μ Charge moves witha velocity of magnitude 5×10⁵ m/s along the +x axis. It experiences no magnetic force. although there is a magnetic field present. The maximum possible magnetic force that the charge with the given speed could experience has a magnitude of 0.340 N. Find the magnitude and direction of the magnetic field.

Answer 1

The magnitude of the magnetic field is approximately 1.02×10⁻⁴ T in the +x direction.

Step-by-step explanation:

The magnitude of the magnetic force experienced by a charged particle moving in a magnetic field is given by the formula F = qvBsinθ, where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnitude of the magnetic field, and θ is the angle between the velocity and the magnetic field.

In this case, the force experienced by the particle is 0.340 N, the velocity is 5×10⁵ m/s, and the charge is 6.7⋅μ. Since the maximum force is given, we can write the formula as 0.340 = (6.7⋅μ)(5×10⁵)(B)(sinθ). Solving for B, we find that the magnitude of the magnetic field is approximately 1.02×10⁻⁴ T.

To determine the direction of the magnetic field, we need to use the right-hand rule. If the velocity vector is in the +x direction, and the force experienced by the particle is zero, then the magnetic field must also be in the +x direction.