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In Walter Lewin's lecture on Faraday's law, he does a demonstration with a solenoid where he shows that current is induced in a loop that surrounds the solenoid as the magnetic flux is changing. He claims that the induced current is the same regardless of the size of the loop of wire that he uses (since the magnetic flux is the same in both cases). But it seems to me that the smaller loop should have a higher induced current. Faraday's law says that:

$$\oint_C{\mathbf{E} \cdot d\mathbf{l}} = - \frac{d}{dt}\int_S\mathbf{B} \cdot d\mathbf{S}$$

If we imagine a solenoid with radius $R$ producing a changing magnetic field, and surrounding it is a wire of radius $r_1 > R$ forming contour $C_1$ and a larger wire with radius $r_2 > r_1$ forming a contour $C_2$ (all concentric), we can see that for the flat surfaces $S_1$ and $S_2$ which enclose the contours $C_1$ and $C_2$ respectively the change in magnetic flux must be the same:

$$\int_{S_1}\mathbf{B} \cdot d\mathbf{S} = \int_{S_2}\mathbf{B} \cdot d\mathbf{S}$$

Then for any given time $t$, $\oint_{C_1}{\mathbf{E_1} \cdot d\mathbf{l}} = \oint_{C_2}{\mathbf{E_2} \cdot d\mathbf{l}}$. By symmetry, the electric fields will be uniform along their contours so $E_1(2\pi r_1) = E_2(2\pi r_2)$. Since $r_2 > r_1 \implies E_1 > E_2$. So this means that the electric field would be larger in the smaller loop and so the current should also be higher.

It seems to me that if you pay close attention to the video, the current reading for the smaller loop is actually higher.

That said, it's much more probable that Walter Lewin is correct and I made a wrong assumption somewhere. Is my understanding of Faraday's law incorrect?

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His demonstration was not capable to show how the electric current depends on the radius of the circuit; the magnetic field change was an abrupt and short pulse made by hand, which makes the pointer in the ammeter to oscillate too quickly to do any reliable measurement. He probably wanted to stress that the induced emf does not depend on the size of the wire loop but he made a mistake and said that the current does not depend on it. He tried to support his claim by the behaviour of the ammeter, but from the standpoint of the electromagnetic theory, this was an error. In theory it is the emf that is independent of the size of the loop. Larger circuit made with the same wire will have larger Ohmic resistance and should experience lower current.

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  • $\begingroup$ So what happens when you put two wires of the same resistance but each with a different radius? The induced emf should be the same, and according to Ohm's law so should the current. But according to my derivation above, the electric field will be higher with smaller radius, so the current should also be higher with smaller radius. Clearly, there must be an inconsistency somewhere but I can't figure it out. $\endgroup$ – hesson Apr 18 '14 at 20:17
  • $\begingroup$ Electric current density $j$ will be higher in the wire of the smaller circuit according to the Ohm law $\mathbf j =\sigma \mathbf E$ where $\sigma$ is conductivity of the metal. However, electric current is given by the product $jA$ where $A$ is cross section area of the wire. The smaller circuit will have smaller $A$ so that its resistance $R$ can be the same as that of larger circuit. Current $I= emf/R$ depends only on emf and resistance. $\endgroup$ – Ján Lalinský Apr 18 '14 at 21:05
  • $\begingroup$ Yes thanks, I have a much better understanding now. $\endgroup$ – hesson Apr 18 '14 at 21:39

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