Calculate the magnetic field at 2 points (checking my result)

In summary, the student attempted to solve the homework problem using Ampere's law, but was incorrect because the points P_1 and P_2 were not exterior to the wire. The student calculated the magnetic field at points P_1 and P_2 by drawing an Amperian loop passing through each point, and found that the field was positive at P_1 but negative at P_2.
  • #1
fluidistic
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Homework Statement


See the picture for a clear description of the situation. I must find the magnetic field at point [tex]P_1[/tex] and [tex]P_2[/tex]. The picture represent a transversal cut of a wire carrying a current I going off the page to our direction. The density of the current is J.


Homework Equations

None given.



The Attempt at a Solution


I used Ampere's law. I know the figure doesn't seem enough symmetric to use Ampere's law, but I think it is. I'll explain why: I though the problem like if there were 2 wires. One of radius 2a and with a current I in our direction and another wire with radius a with current -I/4 (since the area of the "hole" of the wire is one fourth of the area of what would be a circular wire of radius 2a.)
Thanks to Ampere's law, [tex]B_1=\frac{\mu _0 I}{2 \pi d}[/tex].
[tex]B_1(P_1)=\frac{\mu _0 I}{2 \pi (2a+d)}[/tex] and [tex]B_1 (P_2)= \frac{\mu _0 I}{2 \pi (2a+d)}[/tex].

Now [tex]B_2 =-\frac{\mu _0 I}{8\pi d'}[/tex].
[tex]B_2(P_1)=-\frac{\mu _0 I}{8\pi (a+d)}[/tex] and [tex]B_2(P_2)=-\frac{\mu _0 I}{8 \pi (3a+d)}[/tex].
At last, [tex]B(P_1)=B_1(P_1)+B_2(P_1)[/tex] and [tex]B(P_2)=B_1(P_2)+B_2(P_2)[/tex].


Am I right?
 

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  • #2
fluidistic said:
Thanks to Ampere's law, [tex]B_1=\frac{\mu _0 I}{2 \pi d}[/tex].
[tex]B_1(P_1)=\frac{\mu _0 I}{2 \pi (2a+d)}[/tex] and [tex]B_1 (P_2)= \frac{\mu _0 I}{2 \pi (2a+d)}[/tex].

Your approach is correct, but your use of Ampere's Law is not. It is valid for points external to the wire. You need the field for a point inside the wire. If you draw an Amperian loop passing through that point, the current enclosed by the loop is less than the total current I, is it not? So what does Ampere's Law say in this case?
 
  • #3
kuruman said:
Your approach is correct, but your use of Ampere's Law is not. It is valid for points external to the wire. You need the field for a point inside the wire. If you draw an Amperian loop passing through that point, the current enclosed by the loop is less than the total current I, is it not? So what does Ampere's Law say in this case?

You are absolutely right on the fact that my use of Ampere's law is valid only at exterior points of the wire. In fact [tex]P_1[/tex] and [tex]P_2[/tex] are exterior points.
The picture might be confusing now I realize. There's like a very small circle which says "I". In fact this circle only represent the sense of the current, the point in it means that it's going into our direction and from us to the sheet of paper.

Am I misunderstanding you?
 
  • #4
Sorry, I missed that the two points are external. :blushing:

You need to watch your signs. If B1 is positive at P1, then it must be negative at P2. At each point, B2 must have opposite sign to B1 because the current is in the opposite direction.
 
Last edited:
  • #5
kuruman said:
Yes you are misunderstanding me. To calculate the field inside a solid wire of radius 2a, construct an Amperian loop or radius r (r<2a) concentric with the wire. Ampere's Law says that the line integral of B around the loop is equal to μ0 times the current enclosed by the loop. I am saying that that enclosed current is not I but only a fraction of I. Naturally, the fraction depends on r - you enclose current I only when r = 2a.

Yes I know, but why would it help me? I don't have to calculate the magnetic field at any point inside the wire.
Anyway, [tex]I=\pi (2a)^2 J \Rightarrow J=\frac{I}{4\pi a^2}[/tex] and because J is constant in all the wire, I', the current enclosed by any wire of radius r is worth [tex]I'=\frac{\pi r^2 I}{4\pi a^2}[/tex]. But I don't see the application. I guess I'm still not understanding you. (I'm sorry about it!)
 
  • #6
I edited my previous post. You did not misunderstand me, I misunderstood you.
 
  • #7
Ok no problem! I thank you for helping me!


kuruman said:
Sorry, I missed that the two points are external. :blushing:

You need to watch your signs. If B1 is positive at P1, then it must be negative at P2. At each point, B2 must have opposite sign to B1 because the current is in the opposite direction.

My [tex]B_2[/tex] has opposed sign to [tex]B_1[/tex] in both [tex]P_1[/tex] and [tex]P_2[/tex]. But I don't see why my [tex]B_1(P_1)[/tex] should be positive and not my [tex]B_1(P_2)[/tex].
 

Related to Calculate the magnetic field at 2 points (checking my result)

1. How do you calculate the magnetic field at a certain point?

To calculate the magnetic field at a certain point, you will need to know the strength and direction of the current, the distance between the point and the current, and the permeability of the medium in which the current is flowing. You can use the formula B = (mu * I) / (2 * pi * r), where B is the magnetic field, mu is the permeability, I is the current, and r is the distance.

2. Why is it important to check the result when calculating the magnetic field at 2 points?

Checking the result is important to ensure accuracy and validity of your calculations. Mistakes can easily occur when dealing with complex equations, and checking the result can help catch any errors that may have been made.

3. What are some common mistakes when calculating the magnetic field at 2 points?

Some common mistakes when calculating the magnetic field at 2 points include using the wrong equation, using incorrect values for current or distance, and forgetting to account for the permeability of the medium. It is important to double-check all inputs and calculations to avoid these mistakes.

4. Can the magnetic field at 2 points be calculated for any type of current?

Yes, the magnetic field at 2 points can be calculated for any type of current, as long as you have the necessary information and use the correct equation. However, the direction and strength of the magnetic field may vary depending on the type of current.

5. How can I verify the accuracy of my calculated magnetic field at 2 points?

One way to verify the accuracy of your calculated magnetic field at 2 points is by using a compass. Place the compass near the points where you calculated the magnetic field and observe if the needle aligns with the direction of the magnetic field. If it does, then your calculation is likely accurate.

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