Magnetic Field due to a current loop

Click For Summary
A current loop with a radius of 3 cm carrying 2.6 A produces a magnetic field that was calculated to be 0.000031 T at a distance of 2 cm along the axis. The formula used for the calculation was confirmed to be correct, but the online homework system rejected the answer. Users suggested checking the required units and significant figures, as these can affect the input format. Ultimately, the issue was resolved by entering the answer in scientific notation, which the system required. This highlights the importance of adhering to specific formatting in computerized homework systems.
snoweangel27
Messages
15
Reaction score
0
[SOLVED] Magnetic Field due to a current loop

Homework Statement



A single loop of wire of radius 3 cm carries a current of 2.6 A. What is the magnitude of B on the axis of the loop at 2 cm?


Homework Equations



B=\mu/(4\pi) * 2\pi(R^2)*I/(z^2+R^2)^(3/2)



The Attempt at a Solution


I = 2.6 A
R = .03 m
z = .02m

I calculate .000031 T
It should just be a simple matter of plugging in numbers, but every time I use this equation it ends up being wrong. If anyone could tell me what I might be doing wrong it would be greatly appreciated.
 
Physics news on Phys.org
Looks OK to me. Why do you think it's wrong?
 
It is computerized homework, so when I put in the answer that I calculated it keeps telling me that it is wrong. I have done the same thing with multiple different distances, and they all come up wrong, so I believe that it is something that I am doing, I am just not sure what
 
Those online systems can be flaky. I don't think you're doing anything wrong. Your equation is correct and your answer is correct.

Things to check: Are you sure they want the answer in Teslas? How many sig figs does the system like?
 
I finally figured it out. It turns out that the system decided it wanted it in scientific notation. Thanks for your help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
2K
Replies
4
Views
2K
Replies
10
Views
2K
Replies
8
Views
1K
Replies
2
Views
1K
Replies
8
Views
2K
Replies
1
Views
1K
  • · Replies 37 ·
2
Replies
37
Views
4K