# Finding the magnitude of the magnetic field of a moving point charge

1. Mar 14, 2012

### OmegaFury

Question: A point charge is moving with speed 2 x 107m/s along the x axis. At t=0, the charge is at x= 0m and the magnitude of the field at x=4m is B0. The magnitude of the magnetic field at x= 4m when t= 0.1μs is:

The equation and my attempt at solving it is in the attachment.
--I also converted amperes to 1C/1s and I figured that theta is 90 degrees and the magnitude of the unit vector r is one, and thus the cross product of vector v and unit vector r is 2 x 10^7 m/s. ( 2 x 10^7 times one times sin(90 degrees).

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Last edited: Mar 14, 2012
2. Mar 14, 2012

### Staff: Mentor

Your equation for the magnetic field at a location given by vector r from the moving point charge is:
$$\vec{B} = \frac{\mu_o}{4\pi}q\frac{\vec{v}\times \vec{r}}{r^2}$$
You weren't given a value for the point charge q (so your use of the elementary charge is not warranted).

The problem seems to indicate that the "test location" where the field is to be evaluated is directly along the path of the moving charge (the charge is moving along the x-axis, the test location is at x=4m and no y-offset is given). What should that tell you about the results of the cross product v x r ?

3. Mar 14, 2012

### OmegaFury

That would make the cross product zero, wouldn't it? But that would make the magnetic field zero. If that's true, why would it change when t= 0.1 microseconds?

4. Mar 14, 2012

### Staff: Mentor

Good question It wouldn't change. Is the problem statement exactly as you've given it?

5. Mar 14, 2012

### OmegaFury

Exactly as given. It should be 0 though at any given time if vector v is on the same axis as vector r. Thanks for pointing that out. There might be a typo in the problem or something. If, perhaps, the field point was on a different axis or the charge was moving along a different axis, then the magnetic field would change over time because the cross product wouldn't be 0, and the distance r to the field point would be changing with time.

Last edited: Mar 14, 2012
6. Mar 14, 2012

### Staff: Mentor

Yup. That's my take on it too.