Calculating Magnetic Force: Particle Angle & Field Strength Explained

[MrDMD83]

When a charged particle moves at an angle of 13° with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 2.4F?

 

[HallsofIvy]

Once again! Did you bother to read the forum rules?

 

[kyle8921]

The angle at which a charged particle experiences a magnetic force is determined by the direction of the magnetic field and the velocity of the particle. In this scenario, the charged particle is initially moving at an angle of 13° with respect to the magnetic field and experiences a magnetic force of magnitude F. To experience a magnetic force of magnitude 2.4F, the particle must change its angle with respect to the field.

Using the formula for calculating magnetic force, F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field, we can rearrange the equation to solve for the new angle, θ2, at which the particle will experience a magnetic force of 2.4F.

2.4F = qvBsinθ2

θ2 = sin^-1(2.4F/qvB)

Substituting the given values, we get:

θ2 = sin^-1(2.4F/F) = sin^-1(2.4) = 76.2°

Therefore, the particle must change its angle to 76.2° with respect to the magnetic field in order to experience a magnetic force of magnitude 2.4F, while maintaining the same speed. This demonstrates the relationship between the angle of the particle and the strength of the magnetic force it experiences, highlighting the importance of understanding both factors in magnetic force calculations.