# Magnetic field of an infinite current sheet : Amperes law

1. Nov 8, 2012

### Nyasha

I was asked to find the magnetic field of an infinite current sheet due to amperes law. Is my attempt to the solution correct ? The final answer is correct, but l am doubtful of how l got there.

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2. Nov 9, 2012

### frogjg2003

1. This is a surface current, so it would be represented by $\vec{K}$ not $\vec{J}$.
2. How did $\int\vec{J} \cdot d\vec{s}$ turn into $\int\vec{J} \cdot d\vec{l}$. $d\vec{l}$ is a line element around the circuit, while $d\vec{s}$ is an area element of the enclosed surface. You can't turn one into the other.
3. It helps to define a coordinate system, this way you can loose $d\vec{l}$ in favor of $\hat{x}dx$, $\hat{y}dy$, or $\hat{z}dz$, and the same for $d/vec{s}$. This is will avoid confusion when trying to determine $\vec{B}\cdot d\vec{l}$. $d/vec{l}$ is pointing in the opposite direction along $C_2$, not that
4. This is the most important part. The way you drew the loop, $\vec{J}\cdot d/vec{s}=0$. What that tells you is that there is no field parallel to the current.
5. The way you should have drawn the loop is so that the loop is perpendicular to the current. Then $\vec{J}\cdot d\vec{s}=Kda$ and $\int\vec{J}\cdot d\vec{s}=Kl$. This tells you that the field is perpendicular to the surface current.
6. You can't assume that the field is parallel to the plane, so the C2 and C3 integrals aren't necessarily 0. What happens is that the fields are the same along both lines, but because they are in opposite directions, they cancel.

I have a few more things to say, but my laptop is running out of power, I'll continue in the morning.

3. Nov 9, 2012

### frogjg2003

7. You say $B_1l=-B_2l$ but offer no explanation. The explanation is that, by the right hand rule, the components of the fields parallel to the plane and perpendicular to the current have opposite signs and symmetry says they have to have equal magnitudes.
8. You have to argue that by symmetry that there is no field perpendicular to the plane because there the uniform current distribution will cancel out everything except the component parallel to the plate and perpendicular to the the current.