# Homework Help: Magnetic Field of a Moving Point charge

1. Mar 17, 2012

### OmegaFury

1. The problem statement, all variables and given/known data
A point charge of q1=3.6nC is moving with speed 4.5 x 107m/s parallel to the y-axis along the line x=3m. The magnetic field produced by this charge at the origin when it is at the point x=3m, y=4m is approximately:

2. Relevant equations
vector B= (magnetic constant/4pi) (q(vector v X unit vector r)/r2)
magnetic constant= 4pi x 10-7 (Tesla x meter)/Ampere

3. The attempt at a solution
The problem here is a curious cross product outcome.
vector v= 4.5 x 107j = <0, 4.5 x 107, 0>
vector r= <3-0, 4-0, 0-0>= <3,4,0>
magnitude of r= sqrt(32+42)= 5
unit vector r= vector r/ magnitude of vector r= <3,4,0>/5

$\vec{v}$X$\hat{r}$= <0, 4.5 x 107, 0> X <3, 4, 0> =

((4.5 x 107(0) - 0(4))i - (0(0) - 0(3))j + (0(4) - 4.5 x 107(3))k)/5 =

(-13.5 x 107j)/5= <0, 0, -13.5 x 107>/5

Wouldn't that imply that the cross product vector is in the -Z direction? However, using right hand rules, shouldn't the cross product vector be in the positive Z direction?

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Last edited: Mar 17, 2012
2. Mar 17, 2012

### Redbelly98

Staff Emeritus
Welcome to PF.

In the Biot Savart Law, the direction of r is from the current element or charge to the point where B is being calculated. So r is <-3,-4,0> in this case.

p.s. I presume the charge is moving in the +y direction?

3. Mar 17, 2012

### OmegaFury

Ahhhhh. So I had it reversed.
Yes, the charge is moving in the +y direction.

So the cross product would now be: <0, 4.5 x 107, 0>m/s X <-3, -4, 0>/5 =
(0--(3/5)(4.5 x 107)) k= +2.7 x 107m/s k

Then vector B= (magnetic constant/ 4pi) (3.6 x 10-9C)(2.7 x 107m/s k) / (25m2)= 3.89 x 10-10 T k. Which is equal to 0.39nT k

Thank you for your assistance :)

Last edited: Mar 17, 2012