Question

A particle with a charge of 34.6 $\mu C$ moves with a speed of 68.4 m/s in the positive $x$ direction. The magnetic field in this region of space has a component of 0.466 T in the positive $y$ direction, and a component of 0.876 T in the positive $z$ direction. What is the magnitude of the magnetic force on the particle?

Answer 1

The magnitude of the magnetic force on the particle is 67.86 N.

Step-by-step explanation:

Given that,

Charge,
`q=34.6\ \mu C=34.6* 10^(-6)\ C`

Speed of particle, v = 68.4 m/s in +x direction

Magnetic field,
`B=(0.466 j+0.876 k)\ T`

We need to find the magnitude of the magnetic force on the particle. The magnetic force is given by :

`F=q(v* B)\\\\F=34.6* 10^(-6)(68.4i+0+0* (0+0.466j+0.876 k))\\\\F=\begin{pmatrix}0&-59.9184&31.8744\end{pmatrix}\\\\F=(-59.9184j+31.8744k)\ N`

The magnitude of magnetic force on the particle is :

`|F|=√((-59.9184)^2+(31.8744)^2) \\\\|F|=67.86\ N`

So, the magnitude of the magnetic force on the particle is 67.86 N.