Deriving equation for the magnetic field of a variable gap magnet

[CentreShifter]

I'm being asked to derive the equation for the magnetic field produced by a variable gap magnet. I've been given a few clues, but thus far have been unable to actually complete the derivation. The final equation should be:

[tex]B_{ext}=\frac{mg}{lR}\left(\frac{d}{i}\right)[/tex]

I'm told I need to use the Biot-Savart law and Newton's first (specifically sum of torque = 0).

I have also gone as far as recognizing that:

[tex]B=\frac{mg}{iLsin\theta}[/tex]

is going to be incorporated. This is from the cross product equation for the magnetic force (the force is balancing a current carrying wire within the gap of the magnet). I suspect that [tex]l[/tex] and [tex]L[/tex] may be the same variable, but I'm not sure. I've also been told to observe that for small angles [tex]sin\theta \approx tan\theta[/tex]. Any help would be appreciated.

 

[jbsteele]

The equation for the magnetic field produced by a variable gap magnet can be derived using Biot-Savart law and Newton's first law (sum of torque = 0). The equation is:B_{ext} = \frac{\mu_0}{4\pi}\int \frac{\mathbf{I}\times \mathbf{dl}}{r^2} where $\mathbf{I}$ is the current flowing through the wire in the gap, $\mathbf{dl}$ is the differential element of the wire, and $r$ is the distance from the wire. We can use the cross product equation for the magnetic force to rewrite this equation as:B_{ext} = \frac{\mu_0}{4\pi}\int \frac{I dl \sin\theta}{r^2} where $\theta$ is the angle between the current and the differential element of the wire. We can also express the integral in terms of the length of the wire, $L$, and the current flowing through it, $i$, as follows:B_{ext} = \frac{\mu_0 i L}{4\pi}\int \frac{\sin\theta}{r^2} dl For small angles, we can use the approximation $\sin\theta \approx \tan\theta$. We can then use the equation for the arc length of a circle, $s=r\theta$, to rewrite the integral as:B_{ext} = \frac{\mu_0 i L}{4\pi}\int \frac{\tan\theta}{r} ds Now we can use Newton's first law to express the integral in terms of the gap size, $d$, and the radius of the magnet, $R$, as follows:B_{ext} = \frac{\mu_0 i L}{4\pi R}\int \frac{\tan\theta}{d} ds Finally, we can use the equation for the arc length of a circle, $s=r\theta$, to simplify the integral as:B_{ext} =

 

[nlightner]

To derive the equation for the magnetic field of a variable gap magnet, we can start by using the Biot-Savart law, which states that the magnetic field at a point is proportional to the current and the distance from the point to the current element. In this case, the current element is the wire carrying current through the magnet.

Let's consider a small section of the wire that is a distance d from the point where we want to find the magnetic field. The current through this section is i, and the length of the section is L. We can calculate the magnetic field produced by this section using the Biot-Savart law:

d\vec{B} = \frac{\mu_0}{4\pi} \frac{i d\vec{l} \times \hat{r}}{r^2}

where \vec{l} is the direction of the current element and \hat{r} is the unit vector pointing from the current element to the point where we want to find the magnetic field.

Since we are interested in the magnetic field at a point on the axis of the magnet, we can assume that the angle between \vec{l} and \hat{r} is small, so we can approximate sin\theta \approx tan\theta. This allows us to write:

d\vec{B} \approx \frac{\mu_0}{4\pi} \frac{i d l \tan\theta}{r^2} \hat{\phi}

where \hat{\phi} is the unit vector pointing in the azimuthal direction.

Now, we can integrate over the entire length of the wire to get the total magnetic field at our point:

\vec{B} = \frac{\mu_0}{4\pi} \int_{0}^{L} \frac{i d l \tan\theta}{r^2} \hat{\phi}

Next, we can use Newton's first law, which states that the sum of torque on a system is equal to zero, to simplify this equation. Since the magnet is in equilibrium, the torque due to the magnetic field must be balanced by the torque due to the weight of the magnet. This leads to the equation:

\frac{\mu_0}{4\pi} \int_{0}^{L} \frac{i d l \tan\theta}{r^2} \hat{\phi} = mgR \hat{z}

where R is