Solve Magnetic Problem: Charge Q w/ Vector Force of B=0.25T

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To solve the problem of a charge Q=1.6 µC moving with a velocity of v=4.0 x 10^5 m/s in a magnetic field B=0.25T along the z-axis, the Lorentz force equation is applicable. The force experienced by the charge is given by F=q(E+v×B), where E is the electric field and B is the magnetic field. Since the electric field is not specified, it can be assumed to be zero for this scenario. The key point is that the magnetic force will be perpendicular to both the velocity vector and the magnetic field. Understanding this relationship is crucial for calculating the magnetic force acting on the charge.
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Homework Statement



the charge of Q= 1.6 uC is moving along the line x=y with speed v= 4.0*10^5 m/s. What is the vecotr force it experiences due to the magnetic field B= Boz, where Bo = 0.25T?

Homework Equations





The Attempt at a Solution



i know Fb must be perpendicular to B and B is on the z axis. but i don't know what should i do afterwards. please help.
 
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You just need to use the Lorentz force.

\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})

Remember, \vec{E} and \vec{B} are the fields not produced by charge q.
 

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