Solve Ampere's Law: Get Help w/ Figure P30.23

[CartoonKid]

Need Help -- Ampere's Law

http://www.webassign.net/pse/p30-21alt.gif

Figure P30.23 is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is I1 = 1.20 A out of the monitor, and the current in the outer conductor is I2 = 2.88 A into the monitor. Determine the magnitude and direction of the magnetic field at point b.

My working is:
$$B_1 = \frac{\mu_{0}I_1}{2\Pi(3\times10^{-3})}$$
$$B_2 = \frac{\mu_{0}I_2}{2\Pi(1\times10^{-3})}$$
$$B_2 - B_1 = 496\mu T$$
However, my answer is wrong. Somebody please help me.
The hint given after my wrong submission is:
Apply Ampere's Law and consider the currents inside the Amperian loop only.
Can somebody explain briefly to me, what is the main idea of Ampere's Law and suggest the approach we can use everytime we deal with Ampere's Law question.
Thank you very much.

 

[Andrew Mason]

My working is:
$$B_1 = \frac{\mu_{0}I_1}{2\Pi(3\times10^{-3})}$$
$$B_2 = \frac{\mu_{0}I_2}{2\Pi(1\times10^{-3})}$$
$$B_2 - B_1 = 496\mu T$$
However, my answer is wrong. Somebody please help me.
The hint given after my wrong submission is:
Apply Ampere's Law and consider the currents inside the Amperian loop only.

You are not interested in B at point a. The I in Ampere's law is "enclosed current". At b, the enclosed current is the vector sum of the enclosed currents, ie. I2 - I1. So:

$$B_b = \frac{\mu_{0}(I_2-I_1)}{2\pi r_b}$$

AM

 

[wais]

Ampere's Law is a fundamental law in electromagnetism that relates the magnetic field around a closed loop to the electric currents passing through the area bound by that loop. It states that the line integral of the magnetic field around a closed loop is equal to the permeability of free space times the total current passing through the loop.

To apply Ampere's Law to this problem, we need to choose an Amperian loop that encloses the point b and contains the currents passing through the coaxial cable. In this case, we can choose a circular loop with a radius r = 3 mm, centered at point b.

Next, we need to consider the currents passing through this loop. Since the inner conductor carries a current out of the monitor, it will contribute a positive current to the loop. On the other hand, the outer conductor carries a current into the monitor, which will contribute a negative current to the loop.

Using the right-hand rule, we can determine that the magnetic field due to the inner conductor will be in the clockwise direction, while the magnetic field due to the outer conductor will be in the counterclockwise direction. Therefore, we can write the equation for Ampere's Law as:

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 (I_1 - I_2)

Plugging in the values for the currents and the permeability of free space, we get:

B \oint dl = 4\pi\times 10^{-7} (1.20 - (-2.88))

Solving for B, we get:

B = 496\mu T

This matches our previous calculation, but with the correct direction. So the magnetic field at point b is 496\mu T, pointing into the page.