Homework Help: Magnitude of a magnetic field a a point

1. Oct 15, 2014

Jstuff

A point charge Q moves on the x-axis in the positive direction with a speed of 450 m/s. A point P is on the y-axis at y = +70 mm. The magnetic field produced at point P, as the charge moves through the origin, is equal to -0.8 μT k^. When the charge is at x = +40 mm, what is the magnitude of the magnetic field at point P? (μ0 = 4π × 10-7 T · m/A)

Hey guys so I am having trouble understanding this problem. Well I actually understand the problem, but I am having trouble applying the formulas.

I understand that to do that problem I need to use the formula for the magnetic field created by a moving point charge. B=(μ_0 q r x v)/(4π r^2) to solve for the charge.
I did this for position one where the degree between the angle will be sin(90)=1 and got -.00124C.
I then applied the equation again to solve for the magnetic field at the new position. My trouble I am having is the cross product between r and v at this locations. I put r into unit vector form and crossed it with v in the positive x direction, but I do not get the right answer.

2. Oct 16, 2014

ehild

The formula for B is not correct. It should be $$\vec B = \frac {\mu_0}{4\pi} Q \frac{\vec v\times\hat r}{r^2}$$
so you have to cross the velocity with the unit vector pointing from the charge to P.
ehild

3. Oct 16, 2014

Jstuff

Okay, so then I am crossing (450i) x (.O4i-.07j)1/.08. Correct?

4. Oct 16, 2014

ehild

If 0.08 in the denominator stands for $\sqrt{65}$ then it is approximately correct.