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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 5.38

This is, of course, a problem about the Biot-Savart law.

A square loop is just 4 finite line segments. The field due to the square loop will be the same as the sum of the individual fields from each line segment.

We know the field of a line segment a distance $s$ from us (see Example 5.5 on page 223):

$$\begin{array}{}\text{(Eq. 5.37, page 223)}& B=\frac{{\mu }_{0}I}{4\pi s}(\sin {\theta }_2-\sin {\theta }_1)\end{array}$$

Where ${\theta }_1$ and ${\theta }_2$ depend on the length and position of us with respective to the line segment:

In our case, the red dot (us) will be aligned with the mid-point of the side of length $w$:

To get the sines of those angles, we need the opposite side and the hypotenuse. The opposite side is of length $w/2$, of course, but the hypotenuse requires a bit more work. Notice that:

So we can find the distance between the red dot and the midpoint of the segment, and then use it to calculate the hypotenuse:

We can now write

$$\begin{array}{rl}\sin \left({\theta }_2\right)& =\frac{\text{opposite}}{\text{hypotenuse}}\\ & =\frac{w/2}{\sqrt{z^2+(w^2/2)}}\end{array}$$

Of course, ${\theta }_1=-{\theta }_2$, so:

$$\begin{array}{rl}\sin \left({\theta }_1\right)& =\sin (-{\theta }_2)\\ & =-\sin \left({\theta }_2\right)\\ & =-\frac{w/2}{\sqrt{z^2+(w^2/2)}}\end{array}$$

To write the $B$ field in terms of $w$ and $z$, it is relevant to note that

$$s=\sqrt{z^2+(w/2{)}^2}$$

(This is the distance from the red dot to the mid-point of the segment of length $w$)

With this in mind, equation 5.37 ($B=\frac{{\mu }_{0}I}{4\pi s}(\sin {\theta }_2-\sin {\theta }_1)$) becomes:

$$\begin{array}{rl}B& =\frac{{\mu }_{0}I}{4\pi s}(\frac{w/2}{\sqrt{z^2+(w^2/2)}}--\frac{w/2}{\sqrt{z^2+(w^2/2)}})\\ & =\frac{{\mu }_{0}I}{4\pi s}(\frac{w/2}{\sqrt{z^2+(w^2/2)}}+\frac{w/2}{\sqrt{z^2+(w^2/2)}})\\ & =\frac{{\mu }_{0}I}{4\pi s}\frac{w}{\sqrt{z^2+(w^2/2)}}\\ & =\frac{{\mu }_{0}I}{4\pi }\frac{w}{\sqrt{z^2+(w/2{)}^2}\sqrt{z^2+(w^2/2)}}\end{array}$$

To get the field of the actual square loop configuration, we simply multiply by $4$ (to get each side) and add a corresponding factor to pick up only the vertical component:

$$\frac{w/2}{\sqrt{z^2+(w/2{)}^2}}$$

We then conclude that

$$\mathbf{B}=\frac{{\mu }_{0}I}{2\pi }\frac{1}{z^2+(w/2{)}^2}\frac{w^2}{\sqrt{z^2+(w^2/2)}}\hat{\mathbf{z}}$$ In the limit $z>>w$, the last 2 factors simplify: $$\mathbf{B}\approx \frac{{\mu }_{0}I}{2\pi }\frac{1}{z^2}\frac{w^2}{z}\hat{\mathbf{z}}=\frac{{\mu }_{0}I{w}^2}{2\pi z^3}\hat{\mathbf{z}}$$

With $m=I{w}^2$, we get:

$$\mathbf{B}\approx \frac{{\mu }_{0}m}{2\pi z^3}\hat{\mathbf{z}}$$

As expected.

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