enter image description here

I want to know the image current and its location which satisfies the boundary condition at the interface.

This problem was originated from the problem 6-33 in Fields and wave electromagnetics, D. Cheng, 2nd Ed.


To solve this problem, the following is what i did.

With equations in magnetostatics and its boundary condition,

$$ \mathbf{B}=\mu\mathbf{H}, \qquad B_1=B_2 = \mu_1H_{1n}=\mu_2H_{2n}, \qquad H_{t1}=H_{t2} $$ In this case, the normal boundary condition becomes $$ H_{1n}=\mu_rH_{2n} $$

From the magnetic flux density which I got from the figure, $\mathbf{B}=\frac{\mu_0I}{2\pi(y^2+d^2)}(y\mathbf{\hat{x}}+x\mathbf{\hat{y}})$, normal and tangential magnetic flux density($\mathbf{B}$) and magnetic field intensity($\mathbf{H}$) at interface adjacent to magnetic medium 1 are $$ (\mathbf{\hat{x}})\quad B_{1n}=\frac{\mu_0Iy}{2\pi(y^2+d^2)}, \qquad H_{1n}=\frac{Iy}{2\pi(y^2+d^2)} $$ $$ (\mathbf{\hat{y}})\quad B_{1t}=\frac{\mu_0Ix}{2\pi(y^2+d^2)}, \qquad H_{1t}=\frac{Ix}{2\pi(y^2+d^2)} $$ with this, we can get $\mathbf{B}$ and $\mathbf{H}$ in magnetic medium 2. $$ \to H_{2n}=\frac{Iy}{\mu_r2\pi(y^2+d^2)} \to B_{2n}=\frac{\mu_0Iy}{2\pi(y^2+d^2)} $$ $$ \to H_{2t}=\frac{Ix}{2\pi(y^2+d^2)} \to B_{2t}=\frac{\mu_0\mu_rIx}{2\pi(y^2+d^2)} $$ Finally, $$ \mathbf{B}_2=\frac{\mu_0I}{2\pi(y^2+d^2)}(y\mathbf{\hat{x}}+\mu_rx\mathbf{\hat{y}}) $$ Is this whole procedure right? From this, I cannot infer image current $I_i=\left(\frac{\mu_r-1}{\mu_r+1}\right)I$ as stated in the problem. What makes to satisfy boundary condition? How do I get the image current?

  • $\begingroup$ There is a derivation in the following paper that you may want to look at: Hammond, P. "Electric and magnetic images." Proceedings of the IEE-Part C: Monographs 107.12 (1960): 306-313. $\endgroup$
    – user87661
    Aug 6 '15 at 23:24
  • $\begingroup$ Yeah, I looked up that page, but it was hard for me to interpret. Anyway, I think that's the only answer I would get for this question. $\endgroup$
    – user65452
    Aug 7 '15 at 23:45

You have taken the field in air to be only due to $I$ and that in the magnetic material to be only due to the image current $I_i$. First you find the sum of the fields due to the current $I$ and $I_i$ at the air boundary and then due the same current $I$ and a different image current $-I_i$ (located at the same place as the current $I$ in air: hence,these will be added) at the magnetic material boundary.

Magnetic field in air

There is a current $I$ in air into the page $\bigotimes$ and an image current $I_i$ also into the page $\bigotimes$

The field due to $I$ is


The field due to $I_i$ is


So the net perpendicular field $B^{\perp}_{air}=\frac{\mu_0(I+I_i)}{2\pi(y^2+d^2)}y\mathbf{\hat{x}}$

and the net parallel field $B^{\parallel}_{air}=\frac{\mu_0(I-I_i)}{2\pi(y^2+d^2)}d\mathbf{\hat{y}}$

Magnetic field in magnetic material

There is a current $I-I_i$ in air into the page $\bigotimes$

The field due to $I-I_i$ is


So the net perpendicular field $B^{\perp}_{mag}=\frac{\mu_0\mu_r(I-I_i)}{2\pi(y^2+d^2)}y\mathbf{\hat{x}}$

and the net parallel field $B^{\parallel}_{mag}=\frac{\mu_0\mu_r(I-I_i)}{2\pi(y^2+d^2)}d\mathbf{\hat{y}}$

The boundary conditions



So we have from (1)$$I+I_i=\mu_r(I-I_i)$$ and from (2)$$I-I_i=I-I_i$$

which gives $$I_i=\frac{(\mu_r-1)}{(\mu_r+1)}I$$


You do not care about the magnetic material. Obviously it does affect the magnetic field produced by current I. So the field you considered at the boundary is wrong.

The field you have to consider in the interface is such by the original current plus such by an image current I1. Not only the one by the original current..... That is one field B1, outside. I1 located where the text says.

The field 2 (inside the magnetic material) is again B by the original current, plus B by a new image current I2, both at the same location, as the text says.

Applying the boundary conditions with these two B's, you get a simple equations system... Involving I (known), I1 and I2 (unknown).... Solve, and that's it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.