Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
If a conductive wire was carrying current $I$, it would create a magnetic field:
$$B=\frac{\mu_0I}{2\pi r}$$
However, if it's not a wire, but instead a plate how would the expression change/adjust to that geometry?
Also, if a conductive plate carrying current is placed in a magnetic field, would the geometry (from a wire to a plate) change the Lorentz force equation as well?
Consider a rectangular Amperian loop instead of a circular one, oriented so the normal is perpendicular to the current.
The current enclosed is equal to the current $I$ times the length of the loop.
$$\mu L K = \int B\cdot dl$$
B is perpendicular to the loop along the vertical axis, so the only current left after the dot product is along the top and bottom of the loop. So the integral becomes
$$\mu L K = 2LB$$ which simplifies to $$B=\frac{\mu K}{2}$$
The magnetic field is only perfectly parallel to the sheet when the sheet is infinite, so this is the magnetic field of an infinite charged sheet. A finite sheet will have a less regular field, and will require integration with the Biot-Savart Law.
If you'd prefer that method, the Biot Savart Law states that
$$B = \frac{{\mu _0 }}{{4\pi }}\int\frac{{Id\ell \times {\bf{\hat r}}}}{{r^2}}$$
It's easy to express the distance from some point above the plane to any point on the plane using $\theta$ and $\phi$ and some cosines, and then to integrate over the whole surface. You've probably done something similar with electrostatics.
For a straight conducting wire, the magnetic field curls around it in concentric circular field lines. In the case of a current carrying plate (of infinite spread), you can consider it as a tightly packed combination of an infinite number of parallel wires all carrying the same current, with negligible cross-section along a particular dimension so that the resulting arrangement is two-dimensional. In such a case also, the flow of current have the same direction as were in the wire. But the difference is that the flow of current is now through a surface.
To obtain an expression for the magnetic field due to an infinite conducting sheet, we assume that the current through each wire is $I$ and there are $n$ wires per unit length.
We have the Ampere's law:
$$\oint \textbf{B}.d\textbf{l}=\mu_0 I_{enc}$$
The line integral is through a rectangular loop surrounding the current sheet whose plane is perpendicular to the plane of sheet. $I_{enc}$ is the current enclosed by the Amperian loop.
As you can imagine the magnetic field lines lie in a plane perpendicular to the direction of current through the wire. So here in our sheet,in the two sides perpendicular to the sheet the direction of magnetic field is perpendicular to the $I_{enc}$
Hence
$$\textbf{B}.d\textbf{l}=0$$
and the only contribution is from the two parallel sides and so we write
$$2\int Bdl=\mu_0 I_{enc}$$
or $$2BL=\mu_0 I_{enc}$$
or $$B=\frac{\mu_0 I_{enc}}{2L}$$ where $I_{enc}=INL$
Hence $$B=\frac{\mu_0 IN}{2}$$
which means the field strength now is not dependent on the distance from the current carrying sheet. The same analogy you can find in the case of electric field due to a charged infinitely long sheet.
In the case of magnetic field due to a conducting wire, the field decreases as a function of distance from the wire. When such wires are combined to form a sheet, the magnetic field is independent of the distance from the sheet. You can imagine the magnetic field extending out to infinity (of course t is forming a closed loop of infinite extent). This is how the geometry varies.
Since the field has changed there will be a slight change due to the difference in field value for the Lorentz force also.
I like the above two answers, I think it's more what you're looking for, but I'll still add my two cents..
You could combine Maxwell's equations (while neglecting the timescales for charges to move to conductor surfaces ($\frac{\partial \mathbf{E}}{\partial t}$)) to get the induction equation
$\frac{\partial \mathbf{B}}{\partial t} + \nabla \times \left(\sigma^{-1} \left[ \nabla \times \frac{\mathbf{B}}{\mu} \right] \right) = \nabla \times (\mathbf{u} \times \mathbf{B}), \qquad \nabla \bullet \mathbf{B} = 0.$
Assuming there's no moving conductor, we may write this as
$\frac{\partial \mathbf{B}}{\partial t} = - \nabla \times \left(\sigma^{-1} \left[ \nabla \times \frac{\mathbf{B}}{\mu} + \mathbf{J}_{external} \right] \right).$
Where $\mathbf{J}_{external}$ is the externally applied current. Assuming conductivity outside the conductor is uniform, the steady state magnetic field outside the conductor will be dictated by
$\nabla^2 \frac{\mathbf{B}}{\mu} = \mathbf{0}.$
The BC for the above equation is $\mathbf{B} = \mathbf{0}$ far away, and the BC at the conductor surface will be non-zero in general. Without going further (which requires a bit more, but doable, work), I find this result useful still because it describes how the magnetic field behaves outside a wire or sheet or any sort of conducting domain.
In summary, the induction equation can help give you some intuition as to how the magnetic field behaves outside a conductor (by a Laplace equation with boundary conditions on the conductor).
