# 物理代写|电动力学代写electromagnetism代考|ELEC3104

## 物理代写|电动力学代写electromagnetism代考|THE BIOT-SAVART LAW

Let us consider the current-carrying conductor shown in Figure 3.3. Oersted showed that this conductor will generate a magnetic field that is coaxial to the wire. So, if we place an imaginary unit north pole at a point $\mathrm{P}$, distance $r$ from a small elemental section of the wire of length $\mathrm{d} l$, the wire will experience a force that will tend to push it to the left. (The plan view in Figure 3.3b shows why this is so.) In addition, there will be an equal and opposite force on the north pole due to the field surrounding the wire.

Let us try to find the magnetic field strength, $\delta H_1$, at the north pole, due to the current element formed by $I$ and $\mathrm{d} l$. As we have just discussed, the current element produces a force on the north pole, and the north pole will produce an equal and opposite force on the current element. If we can find these two forces, and then equate them, we should get an expression for the magnetic field strength generated by the wire.

Let us initially consider the field at $\mathrm{d} l$ due to the imaginary north pole of strength $p_{\mathrm{N}}$. As this north pole is a point source, it emits magnetic flux in a radial direction. Thus, we can write the flux density as
$$\boldsymbol{B}_N=\frac{p_N}{4 \pi r^2} \boldsymbol{r}$$
Direct experimental measurement shows that the force on a current-carrying conductor placed in a magnetic field is given by
$$F=B I l$$
where $B$ is the flux density of the magnetic field in which the wire is placed, $I$ is the current flowing through the wire and $l$ is the length of the wire. (We can intuitively reason that this equation is correct by noting that powerful electric motors require a large electric current and contain a large amount of wire – they are very heavy!)
By combining Equations (3.8) and (3.9), we find that the force on the element $\mathrm{d} l$ due to the field emitted by the north pole is
$$\mathrm{d} F=\frac{p_N}{4 \pi r^2} I \mathrm{~d} l$$

## 物理代写|电动力学代写electromagnetism代考|MAGNETIC FIELD STRENGTH

Let us consider the current-carrying wire shown in Figure $3.6$. We want to find the magnetic field strength at a point $\mathrm{P}$, distance $R$ from the wire. To find $H$ at this point, we will determine the field strength due to a small current element, and then integrate the result over the length of the wire.
By applying the Biot-Savart law, we get
$$\mathbf{d} H=\frac{I \mathrm{~d} z}{4 \pi r^2} \sin \theta \mathrm{A} \mathrm{m}^{-1}$$
acting into the page. Now, to find the total field strength, we need to integrate Equation (3.19) with respect to length. Unfortunately, as we move along the wire, the distance $r$ and the angle $\theta$ will vary. So, we need to do some substitution and manipulation before we can do any integration.

Instead of working with the angle $\theta$, we can simplify the integration if we use the angle $\alpha$ instead. So, with reference to Figure 3.6, we can see that $z=R \tan \alpha$ and so $\mathrm{d} z=R \mathrm{~d} \alpha / \cos ^2 \alpha$. As $\sin \theta=R l r=\cos \alpha$, we get $r=R l \cos \alpha$. Thus, Equation (3.19) becomes
$$\mathbf{d} \boldsymbol{H}=\frac{1}{4 \pi} \frac{R \mathrm{~d} \alpha}{\cos ^2 \alpha} \frac{\cos ^2 \alpha}{R^2} \cos \alpha$$

Now, as we move from $-\infty$ to $+\infty$ the angle $\alpha$ varies from $-\pi / 2$ to $+\pi / 2$. So.
\begin{aligned} \boldsymbol{H} &=\frac{I}{4 \pi R} \int_{-\pi / 2}^{+\pi / 2} \cos ^2 \alpha \mathrm{d} \alpha \ &=\left.\left.\frac{I}{4 \pi R}\right|{-\pi / 2} ^{+\pi / 2} \sin \alpha\right|{-1} \ &=\frac{I}{4 \pi R}(1+1) \end{aligned}
and $\mathrm{SO}$,
$$\boldsymbol{H}=\frac{I}{2 \pi R} \text { into the page }$$

## 物理代写|电动力学代写electromagnetism代考|THE BIOT-SAVART LAW

$$\boldsymbol{B}_N=\frac{p_N}{4 \pi r^2} \boldsymbol{r}$$

$$F=B I l$$

$$\mathrm{d} F=\frac{p_N}{4 \pi r^2} I \mathrm{~d} l$$

## 物理代写|电动力学代写electromagnetism代考|MAGNETIC FIELD STRENGTH

$$\mathbf{d} H=\frac{I \mathrm{~d} z}{4 \pi r^2} \sin \theta \mathrm{Am}^{-1}$$

$$\mathbf{d} \boldsymbol{H}=\frac{1}{4 \pi} \frac{R \mathrm{~d} \alpha}{\cos ^2 \alpha} \frac{\cos ^2 \alpha}{R^2} \cos \alpha$$

$$\boldsymbol{H}=\frac{I}{4 \pi R} \int_{-\pi / 2}^{+\pi / 2} \cos ^2 \alpha \mathrm{d} \alpha \quad=\frac{I}{4 \pi R}\left|-\pi / 2^{+\pi / 2} \sin \alpha\right|-1=\frac{I}{4 \pi R}(1+1)$$

$$\boldsymbol{H}=\frac{I}{2 \pi R} \text { into the page }$$

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