Answer:
The magnitude of the magnetic force on the particle is 67.86 N.
Explanation:
Given that,
Charge, [tex]q=34.6\ \mu C=34.6\times 10^{-6}\ C[/tex]
Speed of particle, v = 68.4 m/s in +x direction
Magnetic field, [tex]B=(0.466 j+0.876 k)\ T[/tex]
We need to find the magnitude of the magnetic force on the particle. The magnetic force is given by :
[tex]F=q(v\times B)\\\\F=34.6\times 10^{-6}(68.4i+0+0\times (0+0.466j+0.876 k))\\\\F=\begin{pmatrix}0&-59.9184&31.8744\end{pmatrix}\\\\F=(-59.9184j+31.8744k)\ N[/tex]
The magnitude of magnetic force on the particle is :
[tex]|F|=\sqrt{(-59.9184)^2+(31.8744)^2} \\\\|F|=67.86\ N[/tex]
So, the magnitude of the magnetic force on the particle is 67.86 N.
