Magnetic field outside a semi-infinite solenoid

[Bling Fizikst]

Homework Statement

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Relevant Equations

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The solenoid can be assumed to be a stack of rings of width $dz$ each where $z:0\to-\infty$ . $$i=nIdz$$ $$\sin\theta=\frac{R-r}{l}$$ $$\cos\theta=\frac{z+x}{l}$$ $$dB=\frac{\mu_{\circ}nIR d\varphi dz}{4\pi l^2}$$ Call $\frac{\mu_{\circ}nIR}{4\pi}=\alpha$ $$ B_y =\int dB\cos\theta , B_x=\int dB\sin\theta$$ Doing the computation gives : $$B_y=\frac{2\pi\alpha}{\sqrt{(R-r)^2+x^2}}$$ Similarly , $$B_x=\frac{2\pi\alpha}{R-r}\left(1-\frac{x}{\sqrt{(R-r)^2+x^2}}\right)$$ But it happens to be wrong , not sure where??

 

[haruspex]

Surely $\theta$ is a function of $\phi$.

 

[anuttarasammyak]

Say Origin O(0,0,0), point of observation $P(x,y,0)$, the point on a current ring which consistus of the solenoid $Q(X,R \cos\phi,R\sin\phi)$ where X<0
$$PQ^2=(x-X)^2+(y-R\cos\phi)^2+(R\sin\phi)^2=(x-X)^2+y^2+R^2-2yR\cos\phi$$
Does it go well with your treatment ? In order to apply Biot Savart law, current vector is proportinal to
$$(0, -\sin\phi, \cos\phi)$$
Its vector product with vector PQ is proportional to
$$\mathbf{e}_x(R \cos 2\phi + y \ \cos \phi)+\mathbf{e_r} (x-X)$$
if my math is good. So we can apply Biot Savart law with integration of $\phi$ and X.

 

[Bling Fizikst]

Surely $\theta$ is a function of $\phi$.

I am absolutely unable to figure out how . Isn't $\varphi$ present in a plane perpendicular to where $\theta$ is?

 

[haruspex]

I am absolutely unable to figure out how . Isn't $\varphi$ present in a plane perpendicular to where $\theta$ is?

It's just Pythagoras.
To avoid confusion, I will take the point of interest as being at z' from the solenoid, instead of at x.
In (x,y,z) coordinates, where the negative z axis is the axis of the cylinder, you have a fixed point at (0,r,z'). Your solenoid element is at $(R\cos(\phi), R\sin(\phi),z)$, where z is negative.
The distance between the two is $h=\sqrt{R^2\cos^2(\phi)+( R\sin(\phi)-r)^2+(z-z')^2}$. The components of the force are proportional to $(R\cos(\phi),R\sin(\phi)-r,z-z')h^{-\frac 32}$.
Edit: correction, I doubled up on the square-rooting:
The components of the force are proportional to $(R\cos(\phi),R\sin(\phi)-r,z-z')h^{-3}$.

 

[kuruman]

I would approach this problem differently and exploit the information that we are looking for the field "at a very small distance $r$ from the axis."

I would first integrate the expression for the on-axis field due to a single current loop to find the on-axis field $B_z(0,z_0)$ outside the solenoid at distance $z_0$ from the end. At an off-axis point but at a very small distance $r$ from the axis, one can expect a non-zero radial component and an essentially unchanged axial component, i.e. $B_z(r,z_0)\approx B_z(0,z_0).$ This gives the axial component of the field at the off-axis point $~(r,z_0)$.

I would then invoke the on-axis divergence relation in cylindrical coordinates $$0=\mathbf {\nabla}\cdot\mathbf B=\frac{1}{r}\left. \frac{\partial(rB_r)}{\partial r}\right |_{z=z_0}+\left. \dfrac{\partial B_z(0,z)}{\partial z}\right |_{z=z_0}$$from which $$d(r B_r)=-\left. \dfrac{\partial B_z(0,z)}{\partial z}\right |_{z=z_0}r~dr.$$ To find the radial component, note that the derivative on the right-hand side can be treated as constant and one can integrate, $$r B_r=-\left. \dfrac{\partial B_z(0,z)}{\partial z}\right |_{z=z_0}\int_0^r r~dr$$to get the radial component of the field at the off-axis point $~(r,z_0)$.

 

[Bling Fizikst]

It's just Pythagoras.
To avoid confusion, I will take the point of interest as being at z' from the solenoid, instead of at x.
In (x,y,z) coordinates, where the negative z axis is the axis of the cylinder, you have a fixed point at (0,r,z'). Your solenoid element is at $(R\cos(\phi), R\sin(\phi),z)$, where z is negative.
The distance between the two is $h=\sqrt{R^2\cos^2(\phi)+( R\sin(\phi)-r)^2+(z-z')^2}$. The components of the force are proportional to $(R\cos(\phi),R\sin(\phi)-r,z-z')h^{-\frac 32}$.

Oh yeah! But now that i realize this . It seems impossible to solve it by integration . I end up with : $$dB=\frac{\mu_{\circ}nIR}{4\pi}\frac{d\varphi\cdot dz}{(z-z')^2+R^2-2rR\sin\varphi}$$

 

[kuruman]

Oh yeah! But now that i realize this . It seems impossible to solve it by integration . I end up with : $$dB=\frac{\mu_{\circ}nIR}{4\pi}\frac{d\varphi\cdot dz}{(z-z')^2+R^2-2rR\sin\varphi}$$

Then consider the method I outlined in post #6.

 

[haruspex]

Oh yeah! But now that i realize this . It seems impossible to solve it by integration . I end up with : $$dB=\frac{\mu_{\circ}nIR}{4\pi}\frac{d\varphi\cdot dz}{(z-z')^2+R^2-2rR\sin\varphi}$$

Shouldn't that denominator quadratic be raised to the power of 3/2?
Don't forget to apply an approximation for small r. And as @kuruman notes, it is the field normal to the axis that is interesting.

 

[Bling Fizikst]

I would approach this problem differently and exploit the information that we are looking for the field "at a very small distance $r$ from the axis."

I would first integrate the expression for the on-axis field due to a single current loop to find the on-axis field $B_z(0,z_0)$ outside the solenoid at distance $z_0$ from the end. At an off-axis point but at a very small distance $r$ from the axis, one can expect a non-zero radial component and an essentially unchanged axial component, i.e. $B_z(r,z_0)\approx B_z(0,z_0).$ This gives the axial component of the field at the off-axis point $~(r,z_0)$.

I would then invoke the on-axis divergence relation in cylindrical coordinates $$0=\mathbf {\nabla}\cdot\mathbf B=\frac{1}{r}\left. \frac{\partial(rB_r)}{\partial r}\right |_{z=z_0}+\left. \dfrac{\partial B_z(0,z)}{\partial z}\right |_{z=z_0}$$from which $$d(r B_r)=-\left. \dfrac{\partial B_z(0,z)}{\partial z}\right |_{z=z_0}r~dr.$$ To find the radial component, note that the derivative on the right-hand side can be treated as constant and one can integrate, $$r B_r=-\left. \dfrac{\partial B_z(0,z)}{\partial z}\right |_{z=z_0}\int_0^r r~dr$$to get the radial component of the field at the off-axis point $~(r,z_0)$.

Beautiful! Extremely sorry for my late response .

Let's say the distance from the opening of the solenoid along its axis is at $x_{\circ}$ instead of general $x$ to avoid confusion . From my first post , $$i=nIdz$$ is flowing through ring of width $dz$ at a distance $z$ beyond the origin .

So , the magnetic field at a general $x$ from the origin should be : $$dB_x= \frac{\mu_{\circ}nIR^2 dz}{2\left(R^2+(x+z)^2\right)^{\frac{3}{2}}}$$ Take $$\lambda = \frac{\mu_{\circ}nIR^2}{2}$$ and $$z+x=t , t: x\to -\infty$$ Now , $$B_x=\int_{x}^{-\infty} \lambda (R^2+t^2)^{-\frac{3}{2}} dt =\gamma\int_{x}^{-\infty} -2R^2 t^{-3} (R^2t^{-2}+1)^{-\frac{3}{2}}\cdot dt $$ where $$\gamma= \lambda\cdot\frac{1}{-2R^2}$$ Now , take $$u=R^2t^{-2}+1$$ and integrate to get $$B_{x_{\circ}}=-2\gamma\left(1-\frac{x_{\circ}}{\sqrt{R^2+x_{\circ}^2}}\right)\cdot dt$$

Now , using $$\nabla\cdot B=\frac{1}{r}\frac{\partial rB_r}{\partial r}+\frac{\partial B_x}{\partial x}\mid_{x=x_{\circ}}=0$$ We get : $$ B_r= \frac{-\gamma rR^2}{(R^2+x_{\circ}^2)^{-\frac{3}{2}}}$$

So , the final answer should be : $$\frac{\mu_{\circ} nI}{2}\left(\left(1-\frac{x_{\circ}}{\sqrt{R^2+x_{\circ}^2}}\right)\hat{x} +\frac{rR^2}{2(R^2+x_{\circ}^2)^{-\frac{3}{2}}}\hat{r}\right)$$

 

[Bling Fizikst]

There is another method discussed in this video using magnetic force :

 

[kuruman]

So , the final answer should be : $$\frac{\mu_{\circ} nI}{2}\left(\left(1-\frac{x_{\circ}}{\sqrt{R^2+x_{\circ}^2}}\right)\hat{x} +\frac{rR^2}{2(R^2+x_{\circ}^2)^{-\frac{3}{2}}}\hat{r}\right)$$

That's the answer I got.

 

[Bling Fizikst]

I just remembered the formula from the video that we can simply use for the magnetic field on the solenoidal axis (which is again derived from integration) : $$ \frac{\mu_{\circ}nI}{2}(\sin\alpha+\sin\beta)$$ where $\alpha,\beta$ are signed values .

In our case : $$\alpha=\frac{\pi}{2}$$ and $$\sin(-\beta)=\frac{x}{\sqrt{R^2+x^2}}$$ and then proceed with $$\nabla\cdot B=0$$ Saves so much time and labour!