Particle 1, with a charge q 1 =3.60 microcoulombs and a speed v 1 =862 m/s, trav
ID: 1775184 • Letter: P
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!
Explanation / Answer
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 .