Answer:
0.00221 N
Explanation:
Given that,
The charge on the particle, [tex]q=34\mu C[/tex]
The speed of the particle, v = 65.8 m/s (+x direction)
Magnetic field, B = 0.545 T (in +y direction) and 0.828 T in the positive z direction.
The magnetic force is given by the formula as follows :
[tex]F=q(v\times B)[/tex]
Substitute all the values,
[tex]F=34\times 10^{-6}\times (65.8i\times (0.545j+0.828 k))\\\\=34\times 10^{-6}\times (65.8i\times 0.545j +65.8i\times 0.828 k)\\\\=34\times 10^{-6}\times(35.86k +(-54.48j))\\\\=34\times 10^{-6}\times \sqrt{35.86^2+54.48^2} \\\\=0.00221\ N[/tex]
So, the magnitude of the magnetic force on the particle is equal to 0.00221 N.
