# Magnetic Force and Field

1. The problem statement, all variables, and given/known data
Assume the Earth’s magnetic field is 52.0 mT northward at 60.08 below the horizontal in Atlanta, Georgia. A tube in a neon sign stretches between two diagonally opposite corners of a shop window—which lies in a north-south vertical plane—and carries current 35.0 mA. The current enters the tube at the bottom south corner of the shop’s window. It exits at the opposite corner, which is 1.40 m farther north and 0.850 m higher up. Between these two points, the glowing tube spells out DONUTS. Determine the total vector magnetic force on the tube
How to solve this equation I just don't know
how to do S letter in DONUTS
how to integrate to find its length

haruspex
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## Homework Equations

This is a good place to start. What vector equations have you been taught relating magnetic field, current and force?

This is a good place to start. What vector equations have you been taught relating magnetic field, current and force?
F=integration of I B sin theta ds

f
This is a good place to start. What vector equations have you been taught relating magnetic field, current and force?
for D O the total force is zero because it is closed

This is a good place to start. What vector equations have you been taught relating magnetic field, current and force?
so now we have to find the force on N U T S
I think for N and T it is easy but U S needs theta and I have a problem thinking of the limits of the theta in the integration

TSny
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F=integration of I B sin theta ds
For the vector force, you would have ##\vec F = \int \left( I \vec {dL} \times \vec B \right)##
Try to show that the integral gives the same result no matter what word is spelled-out by the tube in the neon sign.

• SHOORY
For the vector force, you would have ##\vec F = \int \left( I \vec {dL} \times \vec B \right)##
Try to show that the integral gives the same result no matter what word is spelled-out by the tube in the neon sign.
how is that? I mean isn't that... I don't know
so you mean that we ignore the letters and treat it as a rectangle or something

TSny
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how is that? I mean isn't that... I don't know
so you mean that we ignore the letters and treat it as a rectangle or something
Well, you need to show that you can ignore the particular shapes of the letters.

The idea is to simplify the integral ##\vec F = \int \left( I \vec {dL} \times \vec B \right)## by factoring out any constant factors.

Last edited:
rude man
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To help you a bit with what post 8 suggests, pretend you're a bug crawling along the tube from one end to the other. You can take any path you like, but you're a smart bug so for each tiny step you take you keep track of the distance traveled in the direction of the tube and the distance traveled perpendicular to the tube. You do that algebraically, resolving each tiny move you make into components parallel and perpendicular to the tube. You can travel any way whatsoever but you must end up at the other corner of the window.

When you separately add up all the + and - parallel and perpendicular steps, what do you get for total net travel alongside & perpendicular to the tube?

TSny
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To help you a bit with what post 8 suggests, pretend you're a bug crawling along the tube from one end to the other. You can take any path you like, but you're a smart bug so for each tiny step you take you keep track of the distance traveled in the direction of the tube and the distance traveled perpendicular to the tube.

rm, If you crawl along the tube, then doesn't each tiny step take you in the direction of the tube and never perpendicular to the tube?
Did you mean parallel or perpendicular to the diagonal of the window?

rude man
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Did you mean parallel or perpendicular to the diagonal of the window?
Yes.
You can go forwards and backwards, left and right, along the diagonal, all you want, long as you wind up in the other corner. Of course, your net travel has to be forwards in order for you to get from one diagonal corner to the other. Sorry I wasn't too clear on that.