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

Answer:

0.00221 N

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 :

$F=q(v\times B)$

Substitute all the values,

$F=34\times 10^{-6}\times (65.8i\times (0.545j+0.828 k))\\\\=34\times 10^{-6}\times (65.8i\times 0.545j +65.8i\times 0.828 k)\\\\=34\times 10^{-6}\times(35.86k +(-54.48j))\\\\=34\times 10^{-6}\times \sqrt{35.86^2+54.48^2} \\\\=0.00221\ N$

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