# Magnetic Field from Current Segments

1. Feb 21, 2014

### theshonen8899

1. The problem statement, all variables and given/known data
Find Bz(0,0,z1), the z component of the magnetic field at the point P located at x=y=0,z=z1 from the current I flowing over a short distance dl = |dl|*j located at the point rc = x1*i.

2. Relevant equations

Biot-Savart law:
(μo/4∏)*[I*(dL x r)/(|r|^2)]

3. The attempt at a solution

I concluded that the radius would be √(( x1 )2 + (z1 )2) and submitted:

(μo/4∏)*[I*dL/√(( x1 )2 + (z1 )2)]

which was wrong. Then I remember that the equation called for r^2 not r, so I submitted:

(μo/4∏)*[I*dL/(( x1 )2 + (z1 )2)]

which was also wrong. I thought that the cross product would be even but just in case I tried:

(μo/4∏)*[I*-dL/(( x1 )2 + (z1 )2)]

which was still wrong.

Am I getting the radius wrong? I'm not quite sure what's going on.

2. Feb 21, 2014

### Staff: Mentor

Start by writing out the vectors involved. Your radius magnitude squared is okay as $x_1^2 + z_1^2$, but you'll need the radius vector and dL vector components to perform the cross product. Do the cross product!

3. Feb 22, 2014

### theshonen8899

So the cross product I want is $\vec{dl}$ x $\hat{r}$ and the homework tells me that $\vec{dl}$ x $\vec{r_{x}}$ = $dlx_{1}\hat{k}$. Shouldn't $\vec{dl}$ x $\vec{r_{x}}$ be negative since $\hat{j} \times \hat{i} = -\hat{k}$?

So assuming $\vec{dl}$ x $\vec{r_{x}}$ = $dlx_{1}\hat{k}$, we know $\hat{r} = \frac{\vec{r}}{r}$. Then would the answer be (μo/4∏)*[(I*dL)/(√(( x1 )2 + (z1 )2)^3)] since there's now an extra r on the bottom?

Last edited: Feb 22, 2014
4. Feb 22, 2014

### Staff: Mentor

There shouldn't be an "extra" r on the bottom. You've extracted the z-component of the cross product but then ignored it? The z-component of the cross product contains an x1, so why does it disappear when you write the result? And I'm not sure why you've introduced a unit vector in the r direction; they're asking for the z-component only, not the whole vector (and it wouldn't be in the same direction as the r vector anyways).

So the z-component of the cross product, as you've found, is $dl x_1$ . The square of the magnitude of r is $x_1^2 + z_1^2$. That, along with the current magnitude and constants of the Biot-Savart law should be sufficient to write the result.

5. Feb 22, 2014

### theshonen8899

I forgot to mention that I tried (μo/4∏)*[(I*dL*x1)/(( x1 )2 + (z1 )2)] which wasn't correct.

From what I understand the Biot-Savart law is (μo/4∏)*[I*(dL x r hat)/(|r|^2)]. Am I not suppose to be finding the cross product dl x r hat?

Sorry I'm having so much trouble understanding and visualizing this. Thank you for your patience.

6. Feb 22, 2014

### Staff: Mentor

Okay, I think I owe you an apology there. The cross product does indeed involve the unit vector in the direction of r. That puts another $\sqrt{x_1^2 + z_1^2}$ in the denominator along with the $x_1^2 + z_1^2$ that's already there. That yields $(x_1^2 + z_1^2)^{3/2}$ net. With that and the $x_1$ in the numerator I think you'll be alright.

7. Feb 22, 2014

### theshonen8899

Not at all, you helped me get the correct answer and more importantly I get it now. Thanks so much!