## 物理代写|电磁学代写electromagnetism代考|The Force Law

We define a vector field $\mathbf{B}(\mathrm{x})$ called magnetic induction as the field that describes the force acting on a charge $q$ that moves with velocity $\mathbf{v}$ :
$$\mathbf{F}=\text { const } q(\mathbf{v} \times \mathbf{B})$$
where the value of the eonetant will be introduced later. We can identify the field $\mathbf{B}$ by means of this force, called the Lorentz force.

To generalize these concepts, we introduce the notion of current density. If we have different charges with different velocities, the current density is given by
$$\mathbf{j}(\mathbf{x})=\lim {\Delta \tau \rightarrow 0} \frac{\left(\sum_i q i \mathbf{v}_i\right) \text { in } \Delta \tau}{\Delta \tau}$$ In the special case in which all the particles in $\Delta \tau$ have the same velocities $$\mathbf{j}(\mathbf{x})=\mathbf{v} \lim {\Delta \tau \rightarrow 0} \frac{\left(\sum_i q i\right) \text { in } \Delta \tau}{\Delta \tau}=\mathbf{v} \rho(\mathbf{x})$$
where $\rho(\mathbf{x})=$ charge density.
Let us consider now a line element $d l$ long with a cross section $d S$, and let us assume that the charges contained in this element all move the same way (see Fig. 3.1). We can write
$$\mathbf{j}(\mathbf{x}) d \tau=\mathbf{j}(\mathbf{x}) d l d S$$
and define a vectorial line element as follows:
$$d \mathbf{l}=d l \mathbf{n}$$
Then
$$\mathbf{j}(\mathbf{x}) d \tau=|\mathbf{j}| d S d \mathbf{l}=I d \mathbf{l}$$
where $I=$ total current $=$ amount of charges that cross the plane perpendicular to $\mathrm{dl}$ in the unit time.

## 物理代写|电磁学代写electromagnetism代考|The Electromagnetic System of Units

In this system of units the unit of current is defined in terms of the unit of force:
$$\mathbf{F}{12}=\text { const }^{\prime \prime} I_1 I_2\left[\oint \oint\left(d \mathbf{l}_1 \cdot d \mathbf{l}_2\right) \nabla \frac{1}{r{12}}\right]$$
We derive the current units from Eq. (3.3.1) in which we set const ${ }^{\prime \prime}=1$ :
$$[F]=\left[I^2\right]$$
This definition is analogous to that of the unit of charge in the electrostatic system, which is the charge that exercises a force of 1 dyne on a similar charge at a distance of $1 \mathrm{~cm}$. In the present case,
$$[I]=\text { abampere }=\sqrt{\text { dyne }}$$
or
$$\left[I^2\right]{\mathrm{EMU}}=M L T^{-2}$$ Then $$\left[q^2\right]{\mathrm{EMU}}=\left[I^2\right] T^2=M L$$
On the other hand,
$$\left[q^2\right]{\mathrm{ESU}}=\text { dyne } \mathrm{cm}^2=M L^3 T^{-2}$$ Then $$\frac{\left[q^2\right]{\text {ESU }}}{\left[q^2\right]{\mathrm{EMU}}}=\frac{M L^3 T^{-2}}{M L}=L^2 T^{-2}$$ and $$\frac{[q]{\text {ESU }}}{[q]_{\text {EMU }}}=L T^{-1}=\text { dimension of a velocity }$$
Therefore, the electromagnetic unit of charge is a constant $c$ times larger than the electrostatic unit of charge. This constant has the dimensions of a velocity; it has been determined experimentally and found to be
$$c=3 \times 10^{10} \mathrm{~cm} / \mathrm{s}$$
which is the speed of light.
We find
1 abampere $=10$ ampere
1 abcoulomb $=10$ coulomb
1 abvolt $=10^{-8}$ volt
If we have two loops such that the geometrical factor is equal to 1 , and if a current of $10 \mathrm{~A}$ passes through each of them, they will attract or repel each other with the force of 1 dyne.

# 电磁学代考

## 物理代写|电磁学代写electromagnetism代考|The Force Law

$$\mathbf{F}=\text { const } q(\mathbf{v} \times \mathbf{B})$$

$$\mathbf{j}(\mathbf{x})=\lim \Delta \tau \rightarrow 0 \frac{\left(\sum_i q i \mathbf{v}_i\right) \text { in } \Delta \tau}{\Delta \tau}$$

$$\mathbf{j}(\mathbf{x})=\mathbf{v} \lim \Delta \tau \rightarrow 0 \frac{\left(\sum_i q i\right) \text { in } \Delta \tau}{\Delta \tau}=\mathbf{v} \rho(\mathbf{x})$$

$$\mathbf{j}(\mathbf{x}) d \tau=\mathbf{j}(\mathbf{x}) d l d S$$

$$d \mathbf{l}=d l \mathbf{n}$$

$$\mathbf{j}(\mathbf{x}) d \tau=|\mathbf{j}| d S d \mathbf{l}=I d \mathbf{l}$$

## 物理代写|电磁学代写electromagnetism代考|The Electromagnetic System of Units

$$\mathbf{F} 12=\text { const }^{\prime \prime} I_1 I_2\left[\oint \oint\left(d \mathbf{l}_1 \cdot d \mathbf{l}_2\right) \nabla \frac{1}{r 12}\right]$$

$$[F]=\left[I^2\right]$$

$$[I]=\text { abampere }=\sqrt{\text { dyne }}$$

$$\left[I^2\right] \mathrm{EMU}=M L T^{-2}$$

$$\left[q^2\right] \mathrm{EMU}=\left[I^2\right] T^2=M L$$

$$\left[q^2\right] \mathrm{ESU}=\text { dyne } \mathrm{cm}^2=M L^3 T^{-2}$$

$$\frac{\left[q^2\right] \mathrm{ESU}}{\left[q^2\right] \mathrm{EMU}}=\frac{M L^3 T^{-2}}{M L}=L^2 T^{-2}$$

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