# Varying current inducing E-field: example 7.9 in Griffiths

1. Oct 8, 2014

### Nikitin

1. The problem statement, all variables and given/known data
Look at the attached file.

1) Why does Griffith simply say that the E-field of the amperian loop is parallel to the axis of the wire?

2) And how come $\int \vec{E} \cdot d \vec{l} = -E(s) l$ ? Shouldn't it at least be $E(s) 2l$ ? Why the minus sign and $l$ instead of $2l$ ?

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2. Oct 8, 2014

### BvU

Homework forum has a template. It's compulsory.
1) Lenz's law. Nature wants to counteract.
Or Maxwell: if B is perp to the paper (call that the z direction, x to the right along the axis) $(\nabla \times \vec E)_z = {\partial E_y\over \partial x} - {\partial E_x\over \partial y} = -({\partial \vec B\over \partial t})_z$
Now Ey can't depend on x (symmetry), so E = Ex only.
2) dl points in a different direction for the two s. In your unsharp picture I think I distinguish a little arrow on the lower dashed line ?

3. Oct 8, 2014

### Nikitin

I used the template as far as possible,, all I need to know is how griffiths justified that E-field stuff.

1) Fair enough.

2) Yes, you are right about the arrow. And sorry about the resolution (can you read everything?). dl points to the right there, so the path integral is taken counter-clockwise. But how does this make the E-field integral equal to - E(s)*l ?

Last edited: Oct 8, 2014
4. Oct 8, 2014

### RUber

On the path integral $\int_{path} \vec E dl = \int _{s_0}^s \vec E(s') ds' +\int _{l}^0 \vec E(s) dl'+\int _s^{s_0} \vec E(s') ds' +\int _{0}^l \vec E(s_0) dl'$
The pieces $\int _{l}^0 \vec E(s) dl'+\int _{0}^l \vec E(s_0) dl'=\vec E(s_0)l-\vec E(s)l$,
The pieces $\int _{s_0}^s \vec E(s') ds'+\int _s^{s_0} \vec E(s') ds'=0$ since the field in invariant in x.

5. Oct 9, 2014

### Nikitin

ahh, right. now I see my mistakes! thank you :)

And thanks to BvU too!