Question

In a particular region, there is a uniform magnetic field with a magnitude of 2.00 T. You take a particle with a charge of +6.00 10-6 C, and give it an initial speed of 1.40 105 m/s in this field. Assume that, after you release it, the particle is influenced only by the magnetic field.

(a) Under these conditions, what is the maximum possible magnitude of the magnetic force experienced by the particle?

(b) Under these conditions, what is the mimimum possible magnitude of the magnetic force experienced by the particle?

(c) If the particle experiences a force with a magnitude that is 25.0% of the magnitude of the maximum force, what is the angle between the particle's velocity and the magnetic field? Use an angle between 0° and 90°.

Answer 1

a) 1.68 N b) 0 c) 14.5º

Step-by-step explanation:

a)

• The force on a charge moving in a magnetic field, is a vector perpendicular to the plane defined by the velocity of the charge and the magnetic field.
• The magnitude of the force F is given by the following expression:

`F = q*v*B*sin\theta (1)`

• where q= magnitude of the charge of the particle, v=velocity of the particle, B= magnitude of the magnetic field, and θ= angle between v and B.
• Replacing by the known values in (1) we have:

`F = q*v*B*sin\theta (1) = 6.00e-6C*1.40e5m/s*2T*sin\theta=\\ \\ F= 1.68N*sin\theta`

• The maximum possible value of F happens when sin θ =1
• This means that v and B are perpendicular each other (in the same plane)
• So, Fmax = 1.68 N

b)

• If the maximum possible value of F happens when sin θ = 1, the minimum possible is when sin θ = 0.
• In this case, when v and B are parallel each other, the force is just 0.

c)

• If the force is 0.25 of the maximum possible, (which is when sin θ =1), this means that sin θ = 0.25, as it can be seen below:

`F_(\theta) = q*v*B*sin \theta = 0.25*Fmax\\ F_(\theta) = q*v*B*sin\theta= 0.25 (q*v*B*sin 90) \\ sin \theta = 0.25`

`\theta = sin^(-1) (0.25) =14.5 deg`

• The angle between the velocity v and the magnetic field B is 14.5º.