Why is magnetic field half at the sides of a solenoid?

In summary, the B-field at the centre of a solenoid is given by ##\mu_0nI##, while at the ends, it is ##\frac12\mu_0nI##. This is also true for the sides of the solenoid, as well as for other geometric shapes such as toroids, square loops, and single coil loops. This can be explained by applying Ampere's law to an Amperian loop passing through the middle of the wire, which only contains half the current of the entire wire, resulting in half the B-field at the centre of the coil. Alternatively, Biot-Savart's law can be used to show that the field at the origin from
  • #1
phantomvommand
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Homework Statement
Magnetic field at the sides of the solenoid (ie at the wires) is half that at the centre. Why?
Relevant Equations
Biot Savart Law may be helpful.
For a solenoid, magnetic field at the centre = ##\mu_0nI##.

I see the argument on why at the opening at the ends of the solenoid, the B-field is ##\frac12\mu_0nI##.

Apparently, B-field is ##\frac12 \mu_0nI## at the sides of the solenoid too. (ie at/within the wires that make up the solenoid). I would appreciate an explantion as to why this is so.

Lastly, is B-field always ##\frac12## that at the centre of any geometric shape, such as in toroids, square loops, single coil loop etc? If so, is there a general explanation for why this is so?

Thanks for all the help.
 
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  • #2
According to the rules, you need to show some sort of attempt before we can help. However, I'll say 2 words: Ampere's law.
 
  • #3
Steve4Physics said:
According to the rules, you need to show some sort of attempt before we can help. However, I'll say 2 words: Ampere's law.
Thanks for this. Apologies for leaving out my attempt. The argument I thought of is that the field outside the solenoid’s coils is 0, and the field in the centre is ##\mu_0nI##, and taking the average will give you the result. This argument applies to the end of solenoid case as well, although I know it is not very rigorous.

Ampere’s Law seems to give back the formula for the B field at the centre of the coil... I don’t see the factor of 1/2 appearing.
 
  • #4
phantomvommand said:
The argument I thought of is that the field outside the solenoid’s coils is 0, and the field in the centre is ##\mu_0nI##, and taking the average will give you the result. This argument applies to the end of solenoid case as well, although I know it is not very rigorous.
Hi. For a long solenoid, that's a reasonable qualitative argument for the field inside the wire being B/2.

But it's an incorrect argument for the field at the ends being B/2, because, at
the ends, the field does not drop from B to 0 over some short distance.

I've got to go out in a few minutes but I'm sure someone else will offer help. Can I suggest you do a little reading about how Ampere's law (and 'Amperian loops') can be used to find the field in a solenoid? Once you've got the idea, it's fairly straightforward to find the field inside the wire (remembering the current flows uniformly through the wire's cross sectional area).
 
  • #5
Steve4Physics said:
Hi. For a long solenoid, that's a reasonable qualitative argument for the field inside the wire being B/2.

But it's an incorrect argument for the field at the ends being B/2, because, at
the ends, the field does not drop from B to 0 over some short distance.

I've got to go out in a few minutes but I'm sure someone else will offer help. Can I suggest you do a little reading about how Ampere's law (and 'Amperian loops') can be used to find the field in a solenoid? Once you've got the idea, it's fairly straightforward to find the field inside the wire (remembering the current flows uniformly through the wire's cross sectional area).
Thanks for clarifying that it’s the point inside the wire. The place where I got this claim from just said “at the coils”. I suppose it only applies inside the wire.

Not sure if this is the right argument, it sounds too simple:

The amperian loop passing through the middle of the wire only contains half the current of that which includes the entire wire. Applying ampere’s law to such a loop will give half the B field found at the centre of the coil.
 
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  • #6
phantomvommand said:
Not sure if this is the right argument, it sounds too simple:

The amperian loop passing through the middle of the wire only contains half the current of that which includes the entire wire. Applying ampere’s law to such a loop will give half the B field found at the centre of the coil.
Yes. Well done!

(Hopefully your loop is rectangular with one side through the mddle of the wire-turns and the other 3 sides essentially outside the solenoid.)
 
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  • #7
Biot-Savart's law probably gives a better explanation. You can do the complete integration to compute the field at the on-axis point on the face of a solenoid of semi-infinite length=i.e. it extends from ##z= -\infty ## to ## z=0 ##, or you can see by symmetry with Biot-Savart that the field at the origin from a full solenoid from ## z=-\infty ## to ## z=+\infty ## will be twice the value of the semi-infinite solenoid.
 
  • #8
Charles Link said:
Biot-Savart's law probably gives a better explanation. You can do the complete integration to compute the field at the on-axis point on the face of a solenoid of semi-infinite length=i.e. it extends from ##z= -\infty ## to ## z=0 ##, or you can see by symmetry with Biot-Savart that the field at the origin from a full solenoid from ## z=-\infty ## to ## z=+\infty ## will be twice the value of the semi-infinite solenoid.
The original question is about the field in the sides, i.e. inside the conductive windings. Biot-Savart probably can't compete with Ampere in this situation!
 
  • #9
Steve4Physics said:
The original question is about the field in the sides, i.e. inside the conductive windings. Biot-Savart probably can't compete with Ampere in this situation!
Sorry, I didn't read the question carefully enough. Trying to analyze the field inside the conductive windings is a new one to me. For a long solenoid, it will go from ## B_z =n \mu_o I ## from anywhere inside the solenoid to zero outside. Trying to get a value for the ## B ## in the space covering the diameter of the wire looks to me upon reading the post more carefully what the OP may be asking. Edit: I see the OP answered it in post 5=finding ## B ## at the very center of the wire.
 
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  • #10
phantomvommand said:
Thanks for clarifying that it’s the point inside the wire. The place where I got this claim from just said “at the coils”. I suppose it only applies inside the wire.

Not sure if this is the right argument, it sounds too simple:

The amperian loop passing through the middle of the wire only contains half the current of that which includes the entire wire. Applying ampere’s law to such a loop will give half the B field found at the centre of the coil.
Does not make sense to me.
Your amperian loop does not have any current piercing the loop. In other words, at least a component of the current must be perpendicula r to your loop. It isn't.
The current just follows the loop geometry.
In fact, the B field in the wire itself is zero by the same amperian loop law. This assumes a long solenoid.
 
  • #11
rude man said:
Does not make sense to me.
Your amperian loop does not have any current piercing the loop. In other words, at least a component of the current must be perpendicula r to your loop. It isn't.
The current just follows the loop geometry.
In fact, the B field in the wire itself is zero by the same amperian loop law. This assumes a long solenoid.
Maybe you are considering the wrong loop? The loop is like the one shown here:
https://slideplayer.com/slide/9304779/28/images/3/Magnetic+field+of+a+solenoid.jpg
but shifted up, so that line ab passes through the centres of the wires. Then half the current is within the loop.
 
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  • #12
Steve4Physics said:
Maybe you are considering the wrong loop? The loop is like the one shown here:
https://slideplayer.com/slide/9304779/28/images/3/Magnetic+field+of+a+solenoid.jpg
but shifted up, so that line ab passes through the centres of the wires. Then half the current is within the loop.
I agree! But that loop indicates that the axial B field is maximum at the coils nearest the solenoid's inside, decreasing to zero at the outside.The B field is B/2 at the center. Good work!
 
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FAQ: Why is magnetic field half at the sides of a solenoid?

Why is the magnetic field half at the sides of a solenoid?

The magnetic field is half at the sides of a solenoid because of the geometry of the solenoid. The magnetic field lines inside a solenoid are tightly packed and parallel to each other, creating a strong and uniform magnetic field. However, at the sides of the solenoid, the magnetic field lines start to spread out and become less concentrated, resulting in a weaker magnetic field.

How does the number of turns in a solenoid affect the magnetic field at its sides?

The number of turns in a solenoid directly affects the magnetic field at its sides. As the number of turns increases, the magnetic field becomes stronger and more concentrated inside the solenoid. However, at the sides of the solenoid, the magnetic field still spreads out, resulting in a weaker field compared to the center of the solenoid.

Can the magnetic field at the sides of a solenoid be increased?

Yes, the magnetic field at the sides of a solenoid can be increased by increasing the current flowing through the solenoid. This is because the magnetic field is directly proportional to the current. However, the field will still be weaker at the sides compared to the center of the solenoid due to the geometry of the solenoid.

Why is the magnetic field half and not zero at the sides of a solenoid?

The magnetic field is not zero at the sides of a solenoid because there is still some magnetic field present due to the current flowing through the solenoid. However, as mentioned before, the field is weaker at the sides due to the spreading out of the magnetic field lines.

How does the length of a solenoid affect the magnetic field at its sides?

The length of a solenoid does not have a direct effect on the magnetic field at its sides. However, a longer solenoid will have a larger surface area, resulting in a larger area for the magnetic field lines to spread out and become weaker at the sides. Therefore, a longer solenoid may have a slightly weaker magnetic field at its sides compared to a shorter solenoid.

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