Equation of magnetic field produced by a solenoid

In summary, the conversation discusses the equation for the magnetic field strength produced by a solenoid at a distance r from it. The formula is B(r) = (1/r^3)((some stuff*unit vector in r direction) + (some stuff*unit vector in θ direction)), but the value of μ must be calculated. The magnetic moment, μ, can be found by googling "magnetic dipole" and is defined as μ=NIS. The formulas for the magnetic field strength outside the solenoid are given, but they are only accurate for medium distances or larger. It is recommended to use Biot-Savart to approximate the on-axis magnetic field strength outside the solenoid.
  • #1
kregg34
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0

Homework Statement



For a lab we need the equation of the magnetic field strength produced by a solenoid a distance r away from it (outside of it). I tried looking online and couldn't find one that didn't involve a bunch of cross products or similar. Wonder if anyone knows the formula? I know its of the form B(r) = (1/r^3)((some stuff*unit vector in r direction) + (some stuff*unit vector in θ direction)). The experiment only looked at the field strength in the xy-plane (or in polar the rθ-plane) so I do not need the third unit vector in the B(r) equation.
My professor said we can find it online but no luck.

Homework Equations

The Attempt at a Solution

 
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  • #2
kregg34 said:

Homework Statement



For a lab we need the equation of the magnetic field strength produced by a solenoid a distance r away from it (outside of it). I tried looking online and couldn't find one that didn't involve a bunch of cross products or similar. Wonder if anyone knows the formula? I know its of the form B(r) = (1/r^3)((some stuff*unit vector in r direction) + (some stuff*unit vector in θ direction)). The experiment only looked at the field strength in the xy-plane (or in polar the rθ-plane) so I do not need the third unit vector in the B(r) equation.
My professor said we can find it online but no luck.

Homework Equations

The Attempt at a Solution

Google "magnetic dipole". There will be a constant scaling factor that is determined by the geometry of the solenoid including the number of turns, the radius, and the current.
 
  • #3
Charles Link said:
Google "magnetic dipole". There will be a constant scaling factor that is determined by the geometry of the solenoid including the number of turns, the radius, and the current.
Ok I found two equations for the two components, (y is perpendicular to coil, z is parallel)
B(in y) =(|μ|/r^3)*(3cos(θ)-1)
B(in z) =(3|μ|/r^3)*(sin(θ)cos(θ))
but what is μ and how do I calculate it?
 
  • #4
kregg34 said:
Ok I found two equations for the two components, (y is perpendicular to coil, z is parallel)
B(in y) =(|μ|/r^3)*(3cos(θ)-1)
B(in z) =(3|μ|/r^3)*(sin(θ)cos(θ))
but what is μ and how do I calculate it?
Magnetic moment ## \mu=N \, I \, A/c ## in c.g.s. units where ## A ## is the cross-sectional area of the solenoid.. I'll need to look it up to see what the constants are in M.K.S. units. Incidentally, the magnetic dipole formula only works well for medium distances or larger. At close range the magnetic field of a solenoid obeys other formulas. e.g. the magnetic field inside a long solenoid is nearly uniform and is given by ## B=\mu_o n \, I ## where ## n=N/L ## and points in the z direction (in the direction of the axis of the solenoid). Also, in general, the magnetic field inside a solenoid is usually of much more interest than the magnetic field outside of a solenoid at a large distance from it. If your distance ## r ## meanwhile is measured from one opening of the solenoid and is not large compared to the length of the solenoid, you will see an inverse square law for the fall-off of the magnetic field strength with distance i.e. especially on axis as you move away from the exit aperture. Could you please supply some additional info of the geometry of the problem? It could be quite relevant=what was the length ## L ## of the solenoid, what was the radius "a" of the solenoid , and what range did your distance ## r ## cover and where was it measured from? Also, how many turns ## N ## did the solenoid have and what was the current ## I ## ? editing... I would try googling magnetic dipole one more time. Wikipedia does a good job (M.K.S.) and defines ## m=NIS ## (## m ## instead of ## \mu ##, with ## S ## in place of ## A ##), and their formulas for ## B ## have a ## \mu_o/(4 \pi) ## multiplying expressions similar to what you presented. Also your formulas are incorrect. It should read ## B_z=m(\mu_o/(4 \pi))(3 cos^2(\theta)-1)/|r|^3 ## and ## B_y=m(\mu_o/(4 \pi))3 sin(\theta)cos(\theta)/|r|^3 ##. If your distance ## r ## was measured from the center of the solenoid and was a good deal larger than the length of the solenoid, these formulas should be reasonably good.
 
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  • #5
Note in the above that ## \theta ## is measured from the z-axis. Hopefully the formulas are in good agreement with your data. Additional note is the professor may be trying to get you acquainted with microscopic(atomic-size) magnetic dipoles, and this is an experiment with a macroscopic one.
 
  • #6
Computing off-axis magnetic fields of solenoids outside the solenoid is prohibitively difficult, so I assume you're looking for the on-axis B field outside the solenoid.

A standard way is to use Biot-Savart; you can either sum the contributions of each winding to the observation point or, more elegantly, you can approximate by assuming an arbitrarily large number of windings N of arbitrarily small current Δi such that N Δi = no. of turns x current of the solenoid. The latter approach works better the more number of (equally spaced) turns of the solenoid per unit length.
 
  • #7
rude man said:
Computing off-axis magnetic fields of solenoids outside the solenoid is prohibitively difficult, so I assume you're looking for the on-axis B field outside the solenoid.

A standard way is to use Biot-Savart; you can either sum the contributions of each winding to the observation point or, more elegantly, you can approximate by assuming an arbitrarily large number of windings N of arbitrarily small current Δi such that N Δi = no. of turns x current of the solenoid. The latter approach works better the more number of (equally spaced) turns of the solenoid per unit length.
It appears the instructor doesn't expect them to do an exact calculation, but wants them to use the magnetic dipole formulation which assumes the field from the solenoid can be approximated by one plus magnetic pole at the center of one exit aperture and a minus magnetic pole at the center of the other. With the proper geometry at a large enough distance from the solenoid, this can be a good approximation. It's not exact, but that's apparently what their instructor is looking for. It would be interesting to hear from the OP again to give us additional feedback.
 
  • #8
Charles Link said:
It appears the instructor doesn't expect them to do an exact calculation, but wants them to use the magnetic dipole formulation which assumes the field from the solenoid can be approximated by one plus magnetic pole at the center of one exit aperture and a minus magnetic pole at the center of the other. With the proper geometry at a large enough distance from the solenoid, this can be a good approximation. It's not exact, but that's apparently what their instructor is looking for. It would be interesting to hear from the OP again to give us additional feedback.
OK. How do you visualize how the solenoid is oriented? Axis = x axis, solenoid running from -L/2 to +L/2? Or is L infinite and the distance vector is in the y-z plane?
 
  • #9
rude man said:
OK. How do you visualize how the solenoid is oriented? Axis = x axis, solenoid running from -L/2 to +L/2? Or is L infinite and the distance vector is in the y-z plane?
Solenoid is oriented along the z-axis. ## \theta ## is measured from the z-axis. ## L ## is finite, and the shorter, the better. ## \\ ## Additional note: The computation of the field is exact outside the solenoid if one assumes a plus pole evenly distributed across one aperture and a minus pole evenly distributed across the other aperture with the inverse square law. In any case, the dipole formulas are approximations that are reasonably good. They assume the poles are centered at the center of the aperture and that ## r>>L/2 ##. Under these conditions, the dipole formulas are readily derived.
 
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  • #10
Charles Link said:
Solenoid is oriented along the z-axis. ## \theta ## is measured from the z-axis. ## L ## is finite, and the shorter, the better. ## \\ ## Additional note: The computation of the field is exact outside the solenoid if one assumes a plus pole evenly distributed across one aperture and a minus pole evenly distributed across the other aperture with the inverse square law. In any case, the dipole formulas are approximations that are reasonably good. They assume the poles are centered at the center of the aperture and that ## r>>L/2 ##. Under these conditions, the dipole formulas are readily derived.
OK. Not really intro physics IMO.
 
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  • #11
rude man said:
OK. Not really intro physics IMO.
It looks like a rather interesting laboratory experiment, but the professor should provide the necessary equations and not expect them to google something that is this advanced. Most would probably have little luck in finding it. They would have no idea what to google.
 
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  • #12
Yeah, and I find the idea of using the mag moment dubious. A winding far away counts just as much as a winding close-by, which I guess is why you said 'the shorter the better" in regard to solenoid length.
 
  • #13
rude man said:
Yeah, and I find the idea of using the mag moment dubious. A winding far away counts just as much as a winding close-by, which I guess is why you said 'the shorter the better" in regard to solenoid length.
The magnetic field from a magnetic moment follows an inverse cube formula for the far field, and there's likely to be only a limited range where the field from the solenoid is far in excess of the Earth's magnetic field. In a number of ways, the field inside the field inside the solenoid is of much more interest, but I think the professor might be trying to teach them about atomic magnetic dipoles and perhaps even magnetic surface currents. In Griffith's E&M textbook, Griffith's shows that the net effect (the magnetic field) of a uniform collection (billions) of (aligned) magnetic dipoles is equivalent to having magnetic surface currents around the surface of the magnetic solid. The magnetic surface currents can be used to compute the magnetic field of a permanent magnet. We'll need to get further feedback from the OP to see if this might be the impetus of all this.
 
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Related to Equation of magnetic field produced by a solenoid

1. What is a solenoid?

A solenoid is a cylindrical coil of wire that is used to create a magnetic field. It is typically made of a conducting material, such as copper, and has numerous loops of wire wrapped around a central axis.

2. How is the magnetic field produced by a solenoid calculated?

The magnetic field produced by a solenoid can be calculated using the equation B = μ0 * n * I, where B is the magnetic field strength, μ0 is the permeability constant, n is the number of loops per unit length, and I is the current flowing through the solenoid.

3. What factors affect the strength of the magnetic field produced by a solenoid?

The strength of the magnetic field produced by a solenoid is affected by the number of loops of wire, the current flowing through the solenoid, and the permeability of the material inside the solenoid.

4. Can the direction of the magnetic field produced by a solenoid be changed?

Yes, the direction of the magnetic field produced by a solenoid can be changed by reversing the direction of the current flowing through the solenoid. This will result in the magnetic field lines also changing direction.

5. What is the significance of the equation for the magnetic field produced by a solenoid?

The equation for the magnetic field produced by a solenoid is significant because it allows us to accurately calculate the strength of the magnetic field and understand how it is affected by various factors. This is important in many applications, such as in electromagnets and in the design of electronic devices.

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