A particle with a charge of 31 µC moves with a speed of 62 m/s in the positive x-direction. The magnetic field in this region of space has a component of 0.50 T in the positive y-direction, and a component of 0.61 T in the positive z-direction. What is the magnitude and direction of the magnetic force on the particle?
magnitude:_______mN
degrees:________° (from the positive z-axis in the yz-plane)
This has been bothering me for a while. Any help is appreciated.
B=sqrt{B(y)²+B(z)²} =
=sqrt{0.5²+0.67²} = 0.836 T,
F=qvBsinα,
sinα=1,
F=qvB=31•10⁻⁶•62•0.836=1.6•10⁻³ N.
Vector B makes an angle β with z-axis:
tanβ =B(y)/B(z) =0.5/0.67=0.746,
β=36.7°
Vector F makes the angle 90° with z-axis =>
the desired angle is 90°-36.7° =53.3°