How Does Ampere's Law Apply to Conducting Wires and Cylindrical Shells?

[TwinGemini14]

An infinitely long solid conducting wire of radius a = 2 cm centered on the z-axis carries a current I1 = 5 A out of the screen. The current is uniformly distributed over the cross-section of the wire. Co-axial with the wire is an infinitely long thin cylindrical conducting shell of radius of b = 5 cm that carries a current of I2 = 4 A into the screen.

http://i662.photobucket.com/albums/uu347/TwinGemini14/Ampere.gif

1) At x = 1 cm, y = 0 (within the central wire), the magnetic field points

A) in the +x direction.
B) in the +y direction.
C) in the -y direction.

:: This should be B, in the +y direction. Use the right hand rule. Current is going out of the page and curl fingers to show that at (1,0), it should point 'up'.
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2) In the region a < r < b (between the central wire and the thin shell), the magnitude of the magnetic field

A) is constant.
B) is porportional to r.
C) is proportional to 1/r.

:: I'm not sure how to explain it, but my gut feeling is that it is proportional to r. So B.
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3) At x = 6 cm, y = 0 (outside of the shell), the magnetic field points

A) in the -x direction.
B) in the +y direction.
C) in the -y direction.

:: This is very similar to question 1. The answer should be C. Use the right hand rule and notice that at (6,0), the field points 'down'.
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4) What is the magnitude of the magnetic field at x = 1 cm, y = 0 (within the central wire)?

A) 5 µT
B) 10 µT
C) 15 µT
D) 20 µT
E) 25 µT

:: This one I cannot seem to figure out. HELP NEEDED HERE. I've tried finding the current density at a radius of x=1 and using ampere's law, but cannot seem to come to one of the 5 options. Any advice?

B = (4pi*10^-7)(pi*(0.01^2)) / (2pi * 0.01) = 3.14*10^-8 WRONG!
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5) What is the magnitude of the magnetic field at x = 6 cm, y = 0 (outside of the shell)?

A) 0.78 µT
B) 3.33 µT
C) 4.51 µT
D) 8.20 µT
E) 12.7 µT

:: B = ((4pi*10^-7)*(Ienclosed))/(2pi*r)
B = (4pi*10^-7)*(5-4) / (2pi * 0.06)
B = 3.33*10^-6

ANSWER IS B.
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Can somebody please look over my answers? I cannot seem to figure out number 4. Thanks for the help in advance.

 

[anathema]

Your answers for questions 1, 2, and 3 are all correct. For question 4, the correct answer is B) 10 µT. You are on the right track with using Ampere's Law, but there are a few errors in your calculation. Here is the correct approach:

First, we need to find the current enclosed by a circle of radius x=1 cm. This can be done by finding the area of the circle (pi*(0.01^2)) and multiplying it by the current density, which is given as a current per unit area (5 A / pi*(0.02^2)). This gives us an enclosed current of 0.5 A.

Next, we can use Ampere's Law to find the magnetic field at the point (x=1 cm, y=0). This can be written as:

B = (mu0 * Ienclosed) / (2pi * r)

Plugging in the values we found, we get:

B = (4pi*10^-7 * 0.5) / (2pi * 0.01) = 10 µT

Therefore, the correct answer is B) 10 µT.

For question 5, your approach is correct. The only error is in the calculation of the enclosed current, which should be (4 A - 5 A) = -1 A. Using this value, we get:

B = (4pi*10^-7 * (-1)) / (2pi * 0.06) = -3.33 µT

Since the question asks for the magnitude of the magnetic field, the answer is 3.33 µT, or option B.

Overall, your understanding of the concepts and approach to the problems is good. Just be careful with your calculations and units, and you should be able to get the correct answers.

 

[tomcue]

Your answers for questions 1, 2, and 3 are correct. For question 4, you have correctly used Ampere's law to find the magnetic field within the central wire. However, the current density you have used is for the entire wire, not just the cross-section at x=1 cm. To find the current density at x=1 cm, you can use the fact that the current is uniformly distributed over the cross-section, so the current density would be I1/(pi*(0.01)^2). Plugging this into the Ampere's law equation, you should get a magnetic field of 5*10^-6 T, which is closest to option B.

For question 5, your approach and answer are correct. Good job!