Potential Energy for a Magnetic Dipole
Page Claimed by Thomas Henderlong
The Main Idea
If a magnetic dipole is allowed to pivot freely without any outside forces, then it will match it's alignment with the applied magnetic field. The system favors a state of lower potential so the aligned magnetic dipole is associated with a lower potential energy in the applied magnetic field. Think of how a magnets interact with each other when inched closer to each other. They'll align in a way so that the north and south ends point in the same direction or have the north end of one magnet touching the south end of the other magnet (state of lowest potential).
A Mathematical Model
By calculating how much work it takes to move a magnetic dipole out of alignment, we can determine the increased potential energy for that dipole.
Let's take a look at a rectangular circuit on a horizontal axle in a magnetic field pictured below.
Now let's look at this system from the side.
We can see that the magnetic force acting on the circuit pushes the circuit out horizontally. The magnetic force is trying to make the circuit's magnetic dipole moment, μ, point in the same direction of the magnetic field. We can measure the amount of work it takes to move the circuit from an initial angle of θi to a final angle of θf. The work done by pushing the circuit out of alignment changes the magnetic potential energy of the system.
As we push the circuit to rotate a angle of dθ around it's horizontal axis, the total distance that the force is exerted on one side of the circuit is equal to (h/2)*dθ. Keep in mind though that we are exerted force on both sides of the circuit. In the diagram this is indicated as Force by Us. We will have to determine the total amount of work done by using an integral because the force exerted depends on the value of θ and θ increases as we push the circuit more and more out of alignment.
[math]\displaystyle{ Work = ΔU_m = \int\limits_{θ_i}^{θ_f}\ 2IwBsinθ(\frac{h}{2}dθ) = IwhB \int\limits_{θ_i}^{θ_f}\ sinθ dθ }[/math]
Where [math]\displaystyle{ IwBsinθ }[/math] is the perpendicular force exerted on only one side of the circuit. We are pushing on both sides so we multiply that by 2. [math]\displaystyle{ (\frac{h}{2}dθ) }[/math] is the distance that the force is exerted.
[math]\displaystyle{ ΔU_m = IwhB[-cosθ]_{θ_i}^{θ_f} = -IwhB[cosθ_f - cosθ_i] }[/math]
[math]\displaystyle{ ΔU_m = Δ(-μBcosθ) }[/math] since [math]\displaystyle{ μ = IA = Iwh }[/math]
so the potential energy for a dipole is [math]\displaystyle{ U_m = -μBcosθ }[/math].
Potential energy for a dipole can also be written as the dot product [math]\displaystyle{ U_m = -\overrightarrow{μ} \bullet \overrightarrow{B}\ }[/math].
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