Show me how terms with Q disappear

[faheemahmed6000]

The general Ampere force law equation given by Maxwell is:

According to Maxwell, all terms containing function ##Q(r)## will vanish after closed integration w.r.t. ##s'## because they will get reduced to functions of ##r## and the upper and lower integration limits will be same since the circuit is closed. I can only see how 3rd term vanishes after closed integration w.r.t. ##s'##. I can't see how 2nd and 4th term vanishes after closed integration w.r.t. ##s'## as I can't reduce them to functions of ##r##. Please show how to reduce 2nd and 4th term to functions of ##r## so that they vanishes after closed integration w.r.t. ##s'##.

Any help will be appreciated.

 

[AndyW]

Thank you for raising this question about the Ampere force law equation given by Maxwell. I can understand your confusion and I would be happy to explain how the 2nd and 4th terms in the equation vanish after closed integration with respect to ##s'##.

Firstly, let's take a closer look at the general Ampere force law equation given by Maxwell:

$$
F = \frac{\mu_0}{4\pi} \int_C \frac{I(s') \times (r - r')}{|r - r'|^3} ds' \tag{1}
$$

As you correctly pointed out, the 3rd term vanishes after closed integration with respect to ##s'## because it becomes a function of ##r## and the upper and lower integration limits are the same. Now, let's focus on the 2nd term:

$$
\frac{\mu_0}{4\pi} \int_C \frac{I(s') \times (r' - r)}{|r' - r|^3} ds' \tag{2}
$$

To simplify this term, we can use the vector identity:

$$
\frac{1}{|r' - r|^3} = \nabla \cdot \left(\frac{1}{|r' - r|}\right) \tag{3}
$$

Substituting this identity into equation (2), we get:

$$
\frac{\mu_0}{4\pi} \int_C I(s') \nabla \cdot \left(\frac{1}{|r' - r|}\right) ds' \times (r' - r) \tag{4}
$$

Now, using the divergence theorem, we can rewrite this as:

$$
\frac{\mu_0}{4\pi} \int_S I(s') \left(\frac{1}{|r' - r|}\right) \hat{n} ds' \times (r' - r) \tag{5}
$$

where ##S## is the surface enclosed by the closed integration path ##C## and ##\hat{n}## is the unit normal vector to the surface ##S##. Since the integration path ##C## is closed, the surface ##S## is also closed. Therefore, the integral in equation (5) becomes zero due to the cross product of the