Find the magnetic field at the center of a rectangle

[vande060]

Homework Statement

find |B| at the center
of a rectangular loop of wire of sides a and b carrying current I

Homework Equations

B = uI/2piR

The Attempt at a Solution

- current flowing counterclockwise

- original art work :D

I did the proof for biot savart law as a homework problem, and I got B = uI/2piR for straight wire, but the proof I used was based of positive and negative infinity for bounds of the integral. I was wondering however, if the same formula (B = uI/2piR ) could be used to tackle the above problem like this

B = 2(uI/2pi(b/2)) + 2(uI/2pi(a/2))

of course I can do simplification on my own, but I am curious if this is the correct route?

 

[gneill]

For a straight current carrying length of wire, the differential element for the magnetic field at a point is given by

$$dB = \frac{\mu_o I \vec{dL} \times \hat{r}}{4 \pi R^2}$$

where dL is the differential line element of the wire and $$\widehat{r}$$ is a unit vector in the direction of r, the vector from the point of interest to the line element. R is the magnitude of r.

The cross product will simplify to dLsin(θ), where θ is the angle between dL and $$\widehat{r}$$.

To find B you should integrate over the length of each wire section. Symmetry will allow you to boil this down to a couple of integrals, and if you look carefully, both of them have the same form with just a slight change of which constants are plugged where.

 

[vande060]

For a straight current carrying length of wire, the differential element for the magnetic field at a point is given by

$$dB = \frac{\mu_o I \vec{dL} \times \hat{r}}{4 \pi R^2}$$

where dL is the differential line element of the wire and $$\widehat{r}$$ is a unit vector in the direction of r, the vector from the point of interest to the line element. R is the magnitude of r.

The cross product will simplify to dLsin(θ), where θ is the angle between dL and $$\widehat{r}$$.

To find B you should integrate over the length of each wire section. Symmetry will allow you to boil this down to a couple of integrals, and if you look carefully, both of them have the same form with just a slight change of which constants are plugged where.

Ive got in now thanks