# Magnetic field including long wires and a quarter-circle wire.

• Ascendant78
In summary: Well, anyway, haruspex & I agree. You can't always expect unanimity, that's why what we provide is hints, not solutions on a platter.
Ascendant78

## Homework Equations

All included above

## The Attempt at a Solution

Well, my solution is there in the box at the bottom. I solved for each of the wire sections individually, two half-infinite straight wires and one quarter-wire. My answer at the bottom there is -19μ T (in the k-hat direction). However, the answer in the back of our book is -23.8μ T (same direction of course). I have looked through my equations for 15 minutes and can't see anything wrong with them. Maybe our answer in the book is wrong (it wouldn't be the first time), but I also wouldn't be surprised if I'm making some mistake in my quarter-circle equation. If someone can help me out, I'd appreciate it.

Your work looks good up to the very last step where you plug in your numbers. What value are you using for ##\mu_o##?

First, let me repeat that it's beter to maintain symbols to the very end than to substitute numbers prematurely. One big reason is the ability to verify identity of dimensions in each term in an equation. Another is clarity.

You did not need to do an integration for the bend. The distance from anywhere around the bend to the observation point is the same. I believe you are off by a factor of 2 in computing the contribution of the bend.

Unfortunately that seemingly decreases rather than increases the magnitude of your answer. I could find nothing wrong with how you computed B for the two semi-infinite straight sections except I did not substitute actual numbers.

My answer would be μ0I(1/4πr + 1/4πr + 1/8r).

I get the same answer as you do via ##\frac{\mu_0 I}{2\pi r} + \frac{\mu_0 I}{2 r} \frac 14##

rude man said:
First, let me repeat that it's beter to maintain symbols to the very end than to substitute numbers prematurely. One big reason is the ability to verify identity of dimensions in each term in an equation. Another is clarity.

You did not need to do an integration for the bend. The distance from anywhere around the bend to the observation point is the same. I believe you are off by a factor of 2 in computing the contribution of the bend.

Unfortunately that seemingly decreases rather than increases the magnitude of your answer. I could find nothing wrong with how you computed B for the two semi-infinite straight sections except I did not substitute actual numbers.

My answer would be μ0I(1/4πr + 1/4πr + 1/8r).

Thanks for the feedback. As far as not having to do an integration for the bend, what would the formula be then? The only three formulas I have seen for calculating magnetic fields were the two I used here (one for infinitely-long wires and the integral formula) and Ampere's Law. I'm thinking I might know how from the final version of the integral I used, but want to make sure since you said I may have been off by a factor of 2.

Anyway, since there seems to be a bit of mixed feedback from everyone, I am still a bit lost here.

Ascendant78 said:
Thanks for the feedback. As far as not having to do an integration for the bend, what would the formula be then? The only three formulas I have seen for calculating magnetic fields were the two I used here (one for infinitely-long wires and the integral formula) and Ampere's Law. I'm thinking I might know how from the final version of the integral I used, but want to make sure since you said I may have been off by a factor of 2.

Anyway, since there seems to be a bit of mixed feedback from everyone, I am still a bit lost here.

Well, anyway, haruspex & I agree. You can't always expect unanimity, that's why what we provide is hints, not solutions on a platter.

There is nothing wrong with your basic formula for the bend: B = (μ0I/4π) ∫dl x r/r2. Since r is constant all around the bend you can move it outside the integral, then the formula simplifies to B = -(μ0I/4πr2)(πr/2) k = -μI/8r k since ∫dl x r = -πr/2 k. The πr/2 is the path length of 1/4 circle of radius r.

Ascendant78 said:
My answer at the bottom there is -19μ T (in the k-hat direction). However, the answer in the back of our book is -23.8μ T (same direction of course).

I think I see. You have written your answer as a multiple of the constant ##\mu_0##. But the answer in the book is in terms of ##\mu T## where ##\mu## stands for micro. Try substituting the value of ##\mu_0## into your answer and see if it agrees with the book. (Sorry I didn't notice the subscript "0" on the ##\mu## in your hand written answer.)

## 1. What is a magnetic field?

A magnetic field is a region in space where a magnetic force can be observed. It is created by moving electric charges, such as electrons, and is represented by lines of force that point in the direction of the force.

## 2. How is a magnetic field affected by long wires?

Long wires carrying an electric current can create a magnetic field around them. The strength of the magnetic field is directly proportional to the current in the wire and inversely proportional to the distance from the wire. This means that the closer you are to the wire, the stronger the magnetic field will be.

## 3. How is a magnetic field affected by a quarter-circle wire?

A quarter-circle wire, or a wire shaped like a quarter of a circle, can also create a magnetic field when a current is passed through it. The magnetic field produced by a quarter-circle wire is strongest at the center of the circle and decreases as you move away from the center.

## 4. What is the relationship between a magnetic field and electric current?

Magnetic fields and electric currents are closely related. Whenever there is an electric current, there is a magnetic field and vice versa. This is known as electromagnetic induction and is the basis for many modern technologies, such as electric motors and generators.

## 5. How can magnetic fields be used in everyday life?

Magnetic fields have a wide range of applications in our daily lives. They are used in technologies such as MRI machines, speakers, and credit card readers. They are also used in compasses for navigation and in the Earth's magnetic field to protect us from harmful solar radiation.

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