Solving Aluminum Half Ring Current Density Problem

[BradJohnson]

1. 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.


2. Homework Equations
mainly current density J=I/a


3. 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.

 

[anathema]

Dear James,

Thank you for reaching out for help with your question. I am happy to provide some insight and guidance on how to approach this problem.

Firstly, let's start by defining some variables:
- r = radius (in meters)
- t = thickness (in meters)
- V = voltage (in volts)
- I = current (in amperes)
- J = current density (in amperes per square meter)
- ρ = resistivity (in ohm-meters)

Now, let's look at the formula for current density: J = I/A, where A is the cross-sectional area of the material through which the current is flowing. In this case, we are dealing with a half ring, so we need to consider the cross-sectional area of the curved part of the ring.

To find the cross-sectional area of the half ring, we can use the formula for the area of a circular sector: A = (θ/360)πr^2, where θ is the central angle of the sector. In this case, the central angle is 20 degrees, so we get A = (20/360)π(0.023^2 - 0.018^2) = 0.000001692 m^2.

Next, we can use Ohm's Law to find the current flowing through the half ring. Ohm's Law states that V = IR, where R is the resistance of the material. In this case, the left side is at 47.5 microvolts (0.0000475 V) and the right side is at 0 volts, so the voltage difference is 0.0000475 V. We can use this voltage difference and the known values of r and t to find the resistance R using the formula R = ρ(t/A), where ρ is the resistivity of the material. The resistivity of aluminum is 2.65 x 10^-8 ohm-meters, so we get R = (2.65 x 10^-8)(0.0045/0.000001692) = 7.08 x 10^-5 ohms.

Now, we can use Ohm's Law again to find the current I: I = V/R = (0.0000475)/(7.08 x 10^-5) = 0.671 amperes.

Finally, we can use

 

[m2003]

As a scientist, my first step in solving this problem would be to gather all the necessary information and equations. From the given information, we know that the half ring is made of aluminum and has a thickness of 4.5mm. We also know the inner and outer radius, as well as the voltage at each end of the half ring. The equation for current density is also provided, J=I/a, where I is the current and a is the cross-sectional area.

Next, I would use the given voltage values to calculate the electric field, E, using the equation E=V/d, where V is the voltage and d is the thickness. In this case, d=4.5mm. Then, using the alternate form of current density, J= (1/p)E, where p is the resistivity of aluminum, I would calculate the current density at r=2cm.

To calculate the total current flowing, we can use the equation I=J*A, where A is the cross-sectional area of the half ring. This can be calculated by subtracting the area of the inner circle from the area of the outer circle. Finally, to calculate the total resistance between the two ends of the half ring, we can use the equation R=V/I, where V is the voltage difference between the two ends and I is the total current flowing.

In summary, to solve this problem, we need to use the equations for current density, electric field, and resistance, along with the given information about the half ring, to calculate the current density at r=2cm, the total current flowing, and the total resistance between the two ends of the half ring. I hope this helps and please let me know if you have any further questions.