# Magnetic Field of a point charge

1. Mar 29, 2014

### majormaaz

1. The problem statement, all variables and given/known data
A point charge q = -2.9 μC moves along the z-axis with a velocity v = (+7.3 x 105 m/s) k . At the moment it passes the origin, what are the strength and direction of the magnetic field at the following positions? Express each field vector in Cartesian form.

(a) At position r1 = (2.0 cm, 0 cm, 0 cm)
(b) At position r2 = (0 cm, 4.0 cm, 0 cm).
(c) At position r3 = (0 cm, 0 cm 1.5 cm).
(d) At position r4 = (3.5 cm, 1.5 cm, 0 cm).
(e) At position r5 = (3.0 cm, 0 cm, 1.0 cm)

2. Relevant equations
Bpoint charge = [μ0/4pi] * [qv x sin Θ / r2]

3. The attempt at a solution

I saw that this problem has a negative charge, so I'd have to use the RHR and reverse direction to account for the charge being negative. I also got the fact that the magnetic field at a point along the same axis as the charge's velocity is 0 Teslas.

I ended up with
(a) 0 i + _ j + 0 k
(b) _ i + 0 j + 0 k
(c) 0 i + 0 j + 0 k
(d) _ i + _ j + 0 k
(e) 0 i + _ j + 0 k

However, everytime I used the point charge formula I have, I end up with an incorrect answer. For example, in part (a), with a r = 2 cm = 0.02 m, I plugged that into [μ0/4pi] * [|q|v x sin Θ / r2], and then reversing the sign to account for a negative charge, but it didn't work.
i.e. I calculated that for part (a), a value of 2 cm for r would have a magnetic field of -5.29e-4 T in the j direction, but that's apparently not right.

2. Mar 30, 2014

### Simon Bridge

What makes you think that is not right?

$$\vec B = \frac{\mu_0}{4\pi}\frac{q\vec v\times \vec r}{r^3}$$

For part a: $\vec v = (0,0,v)^t,\;\vec r = (x,0,0)^t$
$$\vec B = \frac{\mu_0q}{4\pi x^3}\left|\begin{matrix}\hat\imath & \hat\jmath & \hat k\\ 0 & 0 & v\\ x & 0 & 0\end{matrix}\right| = \frac{\mu_0 qv}{4\pi x^2}\hat\jmath$$... plug the numbers in and show me your working.

ref: http://maxwell.ucdavis.edu/~electro/magnetic_field/pointcharge.html