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

0.00221 N

Explanation:

Given that,

The charge on the particle,

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 :

Substitute all the values,

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

true is the correct answer

hope i

measuring things 1 litre of water