Question

Thanks in advance, I couldn't find the answers to the following questions:

Q1) A particle (q = –4.0 µC, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a direction that is 50° away from that of the magnetic field. The particle is observed to have a magnetic force with a magnitude of 29 ?N. What is the magnitude of the magnetic field?

Q2) An electron has a velocity of 6.0 × 10⁶ m/s in the positive x-direction at a point where the magnetic field has the components Bx = 3.0 T, By = 1.5 T and Bz = 2.0 T. What is the magnitude of the magnetic force on the electron at this point?

Q3) A proton has a speed 6.9 × 10⁵ m/s enters perpendicularly into a uniform 0.50 T magnetic field region. What is the radius of the resulting path? (proton: m = 1.67 × 10^?27kg, q = 1.6 × 10^?19 C)

Answer 1

1) The magnetic force on the particle is

where theta is the angle between the charge and magnetic field

I am assuming F is 29 N

q = -4 µC= -4 x 10^-6 C

v = 2 km/s = 2000 m/s

Therefore the magnitude of the magnetic field is

2) The magnetic force is

since v is in x direction, the y component of the force is

The z component of the force is

The magnitude of the net force is

3) For the proton to move in a circular path in the magnetic field, its magnetic force must balance the centripetal force

Therefore the radius of the resulting path is