In Fig 29-59, length a is$\mathbf{4}\mathbf{.}\mathbf{7}\mathit{c}\mathit{m}$(short) and current $\mathit{i}$is$\mathbf{13}\mathit{A}$. (a) What is the magnitude (into or out of the page) of the magnetic field at point P? (b) What is the direction (into or out of the page) of the magnetic field at point P?

1. The length a is,$4.7 \text{cm}=4.7\times {10}^{-2} \text{m}$.
2. The current is, $i=13 \text{A}$.

The magnetic field at a point due to a current-carrying wire depends upon the distance of the point from the wire and the magnitude of the current in the wire. It is determined using the Biot-Savart law. The direction of the magnetic field is decided by the right hand rule. The net magnetic field at the point is the vector sum of the magnetic fields due to all the wires.

Formula:

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{4}\mathbf{\pi }\mathbf{R}}$

We can write the magnetic field using the Biot Savart law as below:

$d\overrightarrow{B}=\frac{{\mu }_0}{4\pi }.\frac{id\overrightarrow{l}\times \hat{r}}{r^2}$

$d\overrightarrow{l}$represents the displacement vector for the length of the wire. Let’s assume theleft bottom point of the loop as origin, upward direction as positive y direction, and horizontal right direction as positive x direction. In +y direction,$dl$will vary from 0 to$2a.$Therefore, we can write

$d\overrightarrow{l}=dy\hat{j}$

Now to find the$\hat{r},$we need to find$\overrightarrow{r}$ and its magnitude. $\overrightarrow{r}$ is the position vector between $dl$ and the point under consideration. The position vector can be written as

$\overrightarrow{r}=\left(2a\right)\hat{i}+\left(2a-y\right)\hat{j}$

We can writethemagnitude of this as

$r=\sqrt{{\left(2a\right)}^2+{\left(2a-y\right)}^2}$

Using this to write the unit vector $\hat{r},$we get

$\hat{r}=\frac{\left(2a\right)\hat{i}+\left(2a-y\right)\hat{j}}{\sqrt{{\left(2a\right)}^2+{\left(2a-y\right)}^2}}$

We can use this in the Biot Savart law equation. So we get

$d\overrightarrow{B}=\frac{{\mu }_0}{4\pi }.\frac{i}{r^2}\left(d\overrightarrow{l}\times \hat{r}\right)$

$$\begin{gathered} \overrightarrow{B}=\frac{{\mu }_0}{4\pi }.\frac{i}{{\left(2a\right)}^2+{\left(2a-y\right)}^2}\left(dy\hat{j}\times \frac{\left(2a\right)\hat{i}+\left(2a-y\right)\hat{j}}{\sqrt{{\left(2a\right)}^2+{\left(2a-y\right)}^2}}\right) \\ d\overrightarrow{B}=\frac{{\mu }_0}{4\pi }.\frac{-i2ady\hat{k}}{{[{\left(2a\right)}^2+\left(2a-y\right)]}^{3/2}} \end{gathered}$$

To find the magnetic field due to the long wire, we can integrate the above equation between the limits$0$ and$2a$ .

$$\begin{gathered} \overrightarrow{B_1}={\int }_0^{2a}d\overrightarrow{B} \\ \overrightarrow{B_1}={\int }_0^{2a}\frac{{\mu }_0}{4\mathrm{\pi }}·\frac{-i2ady\hat{k}}{{\left[{\left(2a\right)}^2+\left(2a-y\right)\right]}^{3/2}} \\ {\overrightarrow{B}}_1=\frac{-2ai{\mu }_0}{4\mathrm{\pi }}·{\int }_0^{2a}\frac{dy\hat{k}}{{\left[{\left(2a\right)}^2+\left(2a-y\right)\right]}^{3/2}} \end{gathered}$$

Integrating this, we get

${\overrightarrow{B}}_1=\frac{-2ai{\mu }_0}{4\pi }.{\left[\frac{\left(2a-y\right)}{\left(4a^2\right)\sqrt{4a^2+{\left(2a-y\right)}^2}}\right]}_0^{2a}\hat{k}$

Simplifying this, we have

${\overrightarrow{B}}_1=\frac{{\mu }_{0}i}{8\sqrt{2}.\pi .a}\hat{k}$

Similarly, for shorter wire, we can write

${\overrightarrow{B}}_2=-\frac{{\mu }_{0}i}{4\sqrt{2}.\pi .a}\hat{k}$

The negative sign here is because of the opposite direction of the current in the shorter wire.

So the total field is

$$\begin{gathered} \overrightarrow{B}={\overrightarrow{B}}_1+{\overrightarrow{B}}_2 \\ \overrightarrow{B}=2\left(\frac{{\mu }_{0}i}{8\sqrt{2}.\pi .a}\hat{k}\right)+2\left(-\frac{{\mu }_{0}i}{4\sqrt{2}.\pi .a}\hat{k}\right) \end{gathered}$$

We have multiplied it by 2 because there are 2 long and 2 short wires in the loop.

Simplifying this, we get

$\overrightarrow{B}=\left(\frac{-{\mu }_{0}i}{4\sqrt{2}.\pi .a}\right)\hat{k}$

Now substituting the given values,

$\overrightarrow{B}=\left(\frac{-\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}\right)\times 13A}{4\sqrt{2}\times \mathrm{\pi }\times 4.7\times {10}^{-2}\mathrm{m}}\right)\hat{k}T$

Calculating this, we get

$\overrightarrow{B}=\left(-1.96\times {10}^{-5} \text{T}\right)\hat{k}\approx \left(-2.0\times {10}^{-5} \text{T}\right)\hat{k}$

Therefore, the magnitude of the net field at the given point is $2.0\times {10}^{-5} \text{T}$.

From the above value of $\overrightarrow{B}$, we can see that the direction of the field is along $-\hat{k}$. It implies that it is into the plane of the page.