Question

Particle 1, with a charge q₁=3.60 microcoulombs and a speed v₁=862 m/s, travels at right angles to a uniform magnetic field. The magnetic force it experiences is 4.25 x 10^-3 N. Particle 2, with a charge q₂=53.0 microcoulombs and a speed v₂= 1.30 x 10³ m/s, moves at an angle of 55.0 degrees relative to the same magnetic field. At what angle relative to the magnetic field must particle 2 move if the magnetic force it experiences is to be 0.0500 N?

The answer was given, 32 degrees. I just need to know the formulas and steps taken to find this answer please!

Answer 1

Given that the particle 1 experiences magnetic force F = 4.25 * 10^-3 N

B = magnetic field

Particle 1 travels at right angles , theta = 90 degrees

F = |q| v B sin(theta)

=> 4.27 *10 ^-3 = (3.60 *10^-6 C )(862 m/s ) (sin 90 )

=> B = 1.37 T is the strength of the magnetic field

At what angle relative to the magnetic field(i.e, B = 1.37 T) must particle 2 move if the magnetic force it experiences is to be 0.0500N ?

F = |q| v B * sin(theta)

0.0500 = 53 *10^-6 C * 1.30 * 10^3 m/s * 1.37 T * sin(theta)

=>theta = 31.98 degrees

= 32 degress .