Magnetic field of a coaxial cable

[andrew410]

The figure below 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 1.86 A out of the page, and the current in the outer conductor is 2.13 A into the page.

FIGURE: http://east.ilrn.com/graphing/bca/user/appletImage?dbid=802444537

(a) Determine the magnitude of the magnetic field at point a.
(b) Determine the magnitude of the magnetic field at point b.

I got part a, which is (mu*I1)/(2*pi*.002). I can't seem to get the right answer for part b. I tried (mu*I2)/(2*pi*.006), where I2 equals the current of the outer conductor. Also, I tried (mu*I3)/(2*pi*.002). All these still give me the wrong answer. What am I doing wrong?

Please help me...Any help would be greatly appreciated! :)

 

[s_a]

For b) you need to take a circular Amperian loop of radius 6mm, and the enclosed current is I2 - I1. So if you apply Ampere's Law, the answer is mu*(I2 - I1)/(2*pi*0.006).

 

[thepatientmental]

(a) To find the magnetic field at point a, we can use the formula B = (mu0*I)/(2*pi*r), where mu0 is the permeability of free space, I is the current, and r is the distance from the current. In this case, mu0 = 4*pi*10^-7 N/A^2, I = 1.86 A, and r = 0.002 m. Plugging these values in, we get B = (4*pi*10^-7*1.86)/(2*pi*0.002) = 5.88*10^-4 T.

(b) To find the magnetic field at point b, we need to consider the currents in both the inner and outer conductors. Since the current in the inner conductor is out of the page, it creates a magnetic field into the page at point b. Similarly, the current in the outer conductor is into the page, which also creates a magnetic field into the page at point b. Therefore, we need to add the magnetic fields from both currents.

Using the same formula as before, the magnetic field from the inner conductor at point b is B1 = (4*pi*10^-7*1.86)/(2*pi*0.002) = 5.88*10^-4 T. The magnetic field from the outer conductor at point b is B2 = (4*pi*10^-7*2.13)/(2*pi*0.006) = 1.77*10^-4 T. Adding these two values together, we get the total magnetic field at point b to be B = B1 + B2 = 5.88*10^-4 T + 1.77*10^-4 T = 7.65*10^-4 T.

So, the magnitude of the magnetic field at point b is 7.65*10^-4 T. It is important to note that the direction of the magnetic field at point b is into the page, as both currents are creating magnetic fields into the page.