How would you determine the magnitude of the magnetic field and the speed of q_2q2?

A charge q_1 = 20.0 µC moves with a speed of 4.00 ✕ 10^3 m/s perpendicular to a uniform magnetic field. The charge experiences a magnetic force of 7.46 ✕ 10^-3 N. A second charge q_2 = 7.00 µC travels at an angle of 40.0° with respect to the same magnetic field and experiences a 1.90 ✕ 10^-3 N force.

1 Answer
Jun 7, 2016

B=0.09325T
v=4.5times10^3 ms^-1

Explanation:

The force, F on a charged particle, q, moving perpendicular to a magnetic field, B is given by:
F=Bqv
Where v= velocity of the particle.

So for q_1 we have:
7.46times 10^-3=B*(20times 10^-6)*(4times10^3)
B=0.09325T

If moving at an angle, theta with respect to the field then the equation needs to be modified as:
F=Bqvsintheta

So for q_2

1.9times 10^-3=0.09325*(7times10^-6).v.sin40
v=4528 ms^-1 =4.5times10^3 ms^-1