Respuesta :

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

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