How would you determine the magnitude of the magnetic field and the speed of #q_2#?

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

Answer:

#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#