- #1
FaraDazed
- 347
- 2
I am preparing for an exam and I am going through a past paper which has solutions given for the questions but I need help understanding how the answer comes about. I suspect it may be just the algebra I don't get, but it may be the physics too.
Wasn't sure if this was the correct forum either so move if needed.
The question asks:
A long non-magnetic cylindrical conductor with inner radius R1 and outer radius R2 carries a current I. Using Ampere’s law and some symmetry arguments, find the magnitude of the magnetic field within the conductor and outside it. Assume that the current I is uniformly distributed across the cross-section of the conductor.
For outside, i.e. r > R2 then I personally got that its simply just
[itex]\oint_c \vec{B} \cdot d\vec{r} = \vec{B} 2 \pi r = \mu_0 I \\
\vec{B} = \frac{\mu_0 I}{2 \pi r}
[/itex]
Which was the correct answer in the solutions. But for R1 < r < R2 I struggle to understand where the final answers comes from as the solution given misses a few small steps.
The solution given is:
[itex]
\oint_c \vec{B} \cdot d \vec{r} = \vec{B} 2 \pi r = \mu_0 \vec{j_0} \pi (r^2 - R_1^2) \\
\vec{B} = \frac{\mu_0 I}{2 \pi r} \cdot \frac{r^2 - R_1^2}{R_2^2 - R_1^2 }
[/itex]
I think I understand the first line, where the ##\pi (r^2 - R_1^2) ## on the very RHS is the area of the cross-section of the contour, but I can't see how the second line follows from that? Any help is much appreciated!
Thanks :)
Wasn't sure if this was the correct forum either so move if needed.
The question asks:
A long non-magnetic cylindrical conductor with inner radius R1 and outer radius R2 carries a current I. Using Ampere’s law and some symmetry arguments, find the magnitude of the magnetic field within the conductor and outside it. Assume that the current I is uniformly distributed across the cross-section of the conductor.
For outside, i.e. r > R2 then I personally got that its simply just
[itex]\oint_c \vec{B} \cdot d\vec{r} = \vec{B} 2 \pi r = \mu_0 I \\
\vec{B} = \frac{\mu_0 I}{2 \pi r}
[/itex]
Which was the correct answer in the solutions. But for R1 < r < R2 I struggle to understand where the final answers comes from as the solution given misses a few small steps.
The solution given is:
[itex]
\oint_c \vec{B} \cdot d \vec{r} = \vec{B} 2 \pi r = \mu_0 \vec{j_0} \pi (r^2 - R_1^2) \\
\vec{B} = \frac{\mu_0 I}{2 \pi r} \cdot \frac{r^2 - R_1^2}{R_2^2 - R_1^2 }
[/itex]
I think I understand the first line, where the ##\pi (r^2 - R_1^2) ## on the very RHS is the area of the cross-section of the contour, but I can't see how the second line follows from that? Any help is much appreciated!
Thanks :)