Answer:
F_B = 0.12N
Explanation:
In order to calculate the magnetic force on the charge, you use the following formula:
[tex]\vec{F_B}=q\vec{v}\ X\ \vec{B}[/tex] (1)
q: charge of the particle = 4μC = 4*10^-6 C
v: speed of the charge = 7.5*10^4 m/s
B: magnitude of the magnetic field = 0.4T
The direction of the motion of the charge is perpendicular to the direction of the magnetic field. Then, the magnitude of the magnetic force is:
[tex]F_B=qvBsin90\°\\\\F_B=(4*10^{-6}C)(7.5*10^4 m/s)(0.4T)=0.12N[/tex]
The magnetic force on the charge is 0.12N