# Walter Lewin's lecture on Faraday's law

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?

• 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. Apr 18, 2014 at 21:05