Question

A particle with a charge of 34.0 $\mu C$ moves with a speed of 65.8 m/s in the positive $x$ direction. The magnetic field in this region of space has a component of 0.545 T in the positive $y$ direction, and a component of 0.828 T in the positive $z$ direction. What is the magnitude of the magnetic force on the particle

Answer 1

0.00221 N

Step-by-step explanation:

Given that,

The charge on the particle,
`q=34\mu C`

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 :

`F=q(v* B)`

Substitute all the values,

`F=34* 10^(-6)* (65.8i* (0.545j+0.828 k))\\\\=34* 10^(-6)* (65.8i* 0.545j +65.8i* 0.828 k)\\\\=34* 10^(-6)*(35.86k +(-54.48j))\\\\=34* 10^(-6)* √(35.86^2+54.48^2) \\\\=0.00221\ N`

So, the magnitude of the magnetic force on the particle is equal to 0.00221 N.