Verification of Ampere-Maxwell Law

[Alex145]

Homework Statement

A current I flows along a wire toward a point charge, causing the charge to increase with time. Consider a spherical surface S centred at the charge, with a tiny hole where the wire is – see figure below. The circumference C of this hole is the boundary of the surface S. Verify that the integral form of Maxwell’s equation

$$\oint_{c} \vec B \cdot d \vec s = \int_{s}( \mu_{o} \vec J + \epsilon_{o} \mu_{o} \frac {\partial \vec E} {\partial t}) \cdot d \vec A $$

I attached a picture of the problem for clarification.

Homework Equations

$$B_{wire} = \frac {\mu_{o} I} {4 \pi R} (cos \theta_{1} - cos \theta_{2})$$R is the radius of the small hole.

$$E = \frac {k_{e} q} {r^2}$$r is the distance to the small hole from the built up charge.

The Attempt at a Solution

Considering that the magnetic field from the current carrying wire travels in a circular path around the wire, $\vec B \cdot d \vec s$ becomes $B ds$. The left hand integral would then just be BC. The time derivative of $\vec E$ becomes $k_{e} I/r^2$ and is parallel to the normal to the area $d \vec A$. Plugging all of this into the above equations leaves

$$\frac {\mu_{o} I} {4 \pi R} C (cos \theta_{1} - cos \theta_{2}) = (\mu_{o} J + \mu_{o} \epsilon_{o} \frac {k_{e} I} {r^2})A$$ The area of the small hole is $A = \pi R^2$.

$$\frac {\mu_{o} I} {4 \pi R} C (cos \theta_{1} - cos \theta_{2}) = \mu_{o} I + \mu_{o} \epsilon_{o} \frac {k_{e} I} {r^2} \pi R^2$$ For a "tiny" hole, the value R is much less than r. So, $\theta_{1} \approx 0$ and $\theta_{2} \approx 180$. We then have

$$\frac {\mu_{o} I} {4 \pi R} C (1-(-1)) = \mu_{o} I + \mu_{o} \epsilon_{o} \frac {k_{e} I} {r^2} \pi R^2$$

$$\frac {\mu_{o} I C} {2 \pi R} = \mu_{o} I + \mu_{o} \epsilon_{o} \frac {k_{e} I} {r^2} \pi R^2$$

$$\frac {\mu_{o} I C} {2 \pi R} = \mu_{o} I + \mu_{o} \epsilon_{o} \frac {I} {4 \pi \epsilon_{o} r^2} \pi R^2$$ The circumference C is just $2 \pi R$ so this becomes

$$\mu_{o} I = \mu_{o} I + \mu_{o} \frac {I} {4 r^2} R^2$$ Once again using the approximation that R is much less than r, we find that the second term $\approx 0$ and that both sides are equal. My question is, is there a way to solve this problem without using the above approximations and assumptions? Or even without the use of a formula for the magnetic field from the wire?

Thanks in advance.

 

[kuruman]

Wait a minute. Your figure shows that $S$ is the surface of the large sphere of radius $R$ minus the area of the small circle of circumference $C$. That's fine. But if that's the case, what is the first integral on the right side, namely $\int_{s} \mu_0 \vec J \cdot d \vec A~?$ Isn't it zero because $\vec J$ is zero everywhere on the surface $S$ that you have chosen? Also, can you explain what $\theta_1$ and $\theta_2$ are?

 

[Alex145]

Yeah I was confused about the surface as I didn't find the question clear on that. If that's the case then I'll have to go over the electric flux integral again. The angles $\theta_{1}$ and $\theta_{2}$ are between the axis of the wire and the vector from the wire pointing to any point on the circumference C. In this case I would get

$$\frac {\mu_{o} I C} {2 \pi R} = \mu_{o} \epsilon_{o} \frac {k_{e} I} {r^2} (4 \pi r^2 - \pi R^2)$$

$$\frac {\mu_{o} I 2 \pi R} {2 \pi R} = \mu_{o} \epsilon_{o} \frac {I} {4 \pi \epsilon_{o} r^2} (4 \pi r^2 - \pi R^2)$$

$$\mu_{o} I = \mu_{o} \frac {I} {4 \pi r^2} (4 \pi r^2 - \pi R^2)$$

$$\mu_{o} I = \mu_{o} {I} (1 - \frac { R^2} {4 r^2})$$ And again saying that R is small compared to r, the second term $\approx 0$. Does this make sense? Is there another way to approach the problem?