Magnetic field due to negative charge

[cuddlesome]

I hope you could help me out with this...

A negative charge q=-7.20mC is moving in a reference frame. When the point charge is at the origin the magnetic field it produces at point (25.o cm,0,0) magnetic field is B= (6.0 x10^(-9)T) j("unit vector") and its speed is 800 m/s . what is the x-.y-. z-components of the velocity of charge??

thanks

 

[Hootenanny]

Well, how do you think we should proceed..?

 

[m2003]

for your question! According to the Lorentz force law, a charged particle moving with a velocity v in a magnetic field B will experience a force F given by F=qvxB. In this case, the negative charge q is moving with a velocity of 800 m/s in a magnetic field B=(6.0x10^-9 T) j, which is directed along the y-axis. Therefore, the force experienced by the charge will be F=qvxB=(-7.20x10^-3 C)(800 m/s)(6.0x10^-9 T) j = -34.56x10^-12 N j.

To find the x, y, and z components of the velocity, we can use the fact that the force is equal to the mass of the particle times its acceleration, F=ma. Since the force is only acting in the y-direction, the acceleration will also be in the y-direction. Therefore, we can set the y-components of the force and acceleration equal to each other: -34.56x10^-12 N = m(-ay).

Solving for ay, we get ay = 34.56x10^-12 N/m.

Now, we can use the equation v=at to find the y-component of the velocity. Since we already know the magnitude of the velocity (800 m/s), we can use that to find the time t it takes for the particle to reach that velocity. v=at, so t=v/a = (800 m/s)/(34.56x10^-12 N/m) = 2.3148x10^-8 s.

Now, we can use this time to find the y-component of the velocity using the equation v=at. v= ayt = (34.56x10^-12 N/m)(2.3148x10^-8 s) = 8.0x10^-4 m/s.

Since the velocity is directed along the y-axis, the x and z components will be zero. Therefore, the velocity of the charge is (0 m/s, 8.0x10^-4 m/s, 0 m/s).

I hope this helps! Let me know if you have any further questions.