Determining Radius from Magnetic Field of a Single-Wire Loop

Click For Summary
The discussion revolves around solving a physics problem involving a single-turn wire loop that generates a magnetic field. The primary challenge is determining the radius and current based on given magnetic field strengths at specific distances. Participants clarify the use of the Biot-Savart Law and discuss the unconventional notation used in the original problem. Suggestions are made to simplify the algebra by rearranging equations and taking roots, which may help in finding the radius more easily. The conversation emphasizes the importance of clear notation and proper application of formulas in solving the problem.
frankifur
Messages
3
Reaction score
3
Homework Statement
A single-turn wire loop produces a magnetic field of 41.2 μT at its center, and 5.15 nT on its axis, at 26.0 cm from the loop center.

a. Find the radius

b. Find the current
Relevant Equations
Biot-Savart Law
So I thought I knew how to do this problem but I've run into some issues that make the algebra feel impossible and I am beginning to feel like I'm taking the wrong approach, I ended up rewriting it in a doc because I was concerned maybe my handwriting was the cause of my error so the work is attached.
 

Attachments

Physics news on Phys.org
frankifur said:
Homework Statement: A single-turn wire loop produces a magnetic field of 41.2 μT at its center, and 5.15 nT on its axis, at 26.0 cm from the loop center.

a. Find the radius

b. Find the current
Relevant Equations: Biot-Savart Law

So I thought I knew how to do this problem but I've run into some issues that make the algebra feel impossible and I am beginning to feel like I'm taking the wrong approach, I ended up rewriting it in a doc because I was concerned maybe my handwriting was the cause of my error so the work is attached.
sin for the axial component? Are you sure?
 
haruspex said:
sin for the axial component? Are you sure?
Looks OK to me. Angle ##\theta##, indicated by an arc in the small upper triangle, is equal to the angle indicated by an arc in the larger triangle. The symbols used by the OP to define the sine as ##R/x## are a bit unconventional.
Screen Shot 2023-04-06 at 7.40.00 AM.png
 
Last edited:
To @frankifur:
Note that $$B_{axis}=\frac{B_{center}R^3}{\left[R^2+z^2 \right]^{3/2}}=\frac{B_{center}\cancel{R^3}}{\cancel{R^3}\left[1+(z/R)^2 \right]^{3/2}}.$$Does this help?
 
You don't need to expand the paranthesis. Just take the cubic root of both sides and you have an eqution in R2. Or, if you rearange it as suggested by Kuruman, move the Bcenter
back to the left hand side and take the root of order 3/2. The field values are given numbers.
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?