Struggling with Homework Equations? Get Help Here!

  • Thread starter Thread starter exitwound
  • Start date Start date
  • Tags Tags
    Homework
AI Thread Summary
The discussion revolves around solving a homework problem related to current density and its integration. The user struggles with finding the correct denominator value in the equation J = i/A and seeks clarification on the integration process. After some back-and-forth, it is clarified that the total current i can be calculated by integrating the current density over the area, leading to the correct expression. The final solution for the total current is derived as i = (J_o * 2π * R^2) / 3. The user expresses gratitude for the assistance received in understanding the integration steps.
exitwound
Messages
291
Reaction score
1

Homework Statement



Capture.JPG


Homework Equations



J=i/a

The Attempt at a Solution



Studying for a test. This sample exam went pretty well. I am having trouble with 3 out of the 20. I can't figure this one out. I can't seem to find how to get a denominator value.

The answer is B.

I know that current density is equal to the current over the area.
J=\frac{i}{A}
J_o(r/R)=\frac{i}{\pi (\frac{r}{R}^2)}

I know I'm doing something wrong. Anyone help?
 
Physics news on Phys.org
You are right at this formula:

<br /> J = \frac{i}{A}<br />

But it is just the special case of this:

i = \int JdA
and dA = 2 * pi * r * dr.
 
Oh. Right.

I still don't know how to get a correct answer out of that.

i = \int J \cdot da

i = \int_0^r \frac{J_or}{R} \cdot 2\pi r dr

pulling out the constants:

i = \frac{J_o2\pi}{R} \int_0^r r^2 dr

i = \frac{J_o2\pi}{R}\frac{r^3}{3}|_0^r

i = \frac{J_o2\pi}{R}\frac{r^3}{3}

Where's the extra r in the numerator coming from?
 
When you set r = R, you will get what you want (because problem asks you to find total current, not current at radius r, i think).
 
Last edited:
Oh. because I'm integrating from a radius of 0 to a radius of R. That makes sense.

<br /> i = \frac{J_o2\pi}{R}\frac{r^3}{3}|_0^R<br />

<br /> i = \frac{J_o2\pi}{R}\frac{R^3}{3}<br />

<br /> i = \frac{J_o2\pi R^2}{3}<br />

Thanks :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top