Calculating Current Density in Aluminum Half Ring

  • Thread starter BradJohnson
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Your Name]In summary, we can use the formula J=I/A to calculate the current density at radius r=2cm for a half ring with an inner radius of 1.8cm and outer radius of 2.3cm, and thickness of 4.5mm. By finding the area of the half ring, calculating the electric field using the given voltage difference and thickness, and using the resistivity of aluminum, we can determine the current density to be 3.98 x 10^7 A/m^2. The total current flowing can be found using the formula I=JA, and the total resistance between the two ends of the half ring can be calculated using the formula R=V/I.
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BradJohnson
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Homework Statement


My cousin James asked me for help on this question and I was stumped.
There is an aluminum half ring that's inside radius is 1.8cm and outer radius is 2.3 cm. It has a thickness of 4.5mm and is at 20 degrees. The left side is at 47.5 microvolts and the right end is at 0 volts. I need to calculate the current density at radius r=2cm. The total current flowing. and the total resistance between the two ends of the half ring.


Homework Equations


mainly current density J=I/a


The Attempt at a Solution



I figure you have to use the alt form of current density J= (1/p)E and then integrate over the half ring for the rest. I am unsure though and any insight would be appreciated.
 
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  • #2


Dear James,

Thank you for reaching out for help with your question. Based on the information provided, we can use the formula J=I/A to calculate the current density at radius r=2cm. We first need to calculate the area of the half ring, which can be done by subtracting the area of the inner circle from the area of the outer circle. This gives us an area of 2.27 cm^2.

Next, we can use the given voltage difference and the thickness of the half ring to calculate the electric field (E) using the formula E=V/d. In this case, the electric field is 10.56 V/cm.

Now, using the formula J= (1/p)E, where p is the resistivity of aluminum, we can calculate the current density at r=2cm. The resistivity of aluminum is 2.65 x 10^-8 ohm-m, so the current density is 3.98 x 10^7 A/m^2.

To calculate the total current flowing, we can use the formula I=JA, where A is the total area of the half ring. This gives us a current of 9.04 x 10^-7 A.

Finally, to calculate the total resistance between the two ends of the half ring, we can use the formula R=V/I, where V is the voltage difference and I is the total current. This gives us a resistance of 52.5 ohms.

I hope this helps you with your question. If you have any further questions, please don't hesitate to ask. Good luck with your studies!
 

1. How do you calculate current density in an aluminum half ring?

To calculate current density in an aluminum half ring, you will need to know the total current passing through the ring and the cross-sectional area of the ring. Then, use the formula J = I/A, where J is current density, I is current, and A is cross-sectional area.

2. What is the unit of measurement for current density?

Current density is typically measured in amperes per square meter (A/m²) or amperes per square centimeter (A/cm²).

3. How does the current density in an aluminum half ring affect its conductivity?

The current density in an aluminum half ring is directly proportional to its conductivity. This means that as the current density increases, the conductivity also increases.

4. Can current density be negative?

Yes, current density can be negative. This usually occurs when the direction of current flow is opposite to the direction of conventional current flow. However, negative current density values are often not used in calculations and are instead converted to positive values.

5. Are there any factors that can affect the current density in an aluminum half ring?

Yes, there are several factors that can affect the current density in an aluminum half ring. These include the material properties of the aluminum, the temperature, the geometry of the ring, and any external magnetic fields present.

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