- #1

Luxucs

- 13

- 3

Hi there! I have what I hope is a relatively straightforward question regarding Faraday's law and motional emf, but its been causing me to scratch my head for quite a while.

Consider the diagram attached to this post (source is linked at the bottom). Assume that all of the wires and the rod are perfect conductors, the magnetic field is uniform and time-independent (pointing into the page), and the velocity of the rod is constant (due to some applied force to the bar keeping it that way - we don't need to worry about that right now). We'll assume the self-inductance of the circuit is negligible. The positive x-direction is defined horizontally to the right in the figure, the positive y-direction is upwards, and the positive z-direction is out of the screen.

Now, as the rod moves to the right, the total emf in this circuit can be written as,

$$\epsilon = \oint ({\vec E + \vec v \times \vec B}) \cdot d \vec l$$

Now, given that the wires are thin and all of the charge stays within the wires, we may also write this as,

$$\epsilon = -\frac {d} {dt} \iint{\vec B \cdot d\vec S} = -\frac {d\Phi_B} {dt}$$

Now, my understanding is that this is

$$\oint {\vec E \cdot d\vec l} = -\iint{\frac {\partial \vec B} {\partial t} \cdot d\vec S}$$

As aforementioned, the magnetic field present is time-independent, so we should be able to conclude that,

$$\epsilon = \oint ({\vec v \times \vec B}) \cdot d \vec l$$

From here it is relatively straightforward to calculate the emf in the circuit. The leg of the circuit that is moving is the only part of the integral that contributes to the emf, and since the magnetic field and velocity are both uniform, we can easily see that,

$$\epsilon = vBd$$

Now, this is going to produce a current whose magnitude is given as,

$$I = \frac {vBd} {R}$$

This is all well and straightforward up to here. Now, this current produces an electric field in the resistor. We can see that this electric field in the resistor is conservative from the differential form of Faraday's law,

$$\nabla \times \vec E = -\frac {\partial \vec B} {\partial t} = \vec 0$$

Since the electric field is conservative, we can just determine it from the gradient of ##\epsilon##, with the potential difference across the resistor being equal in magnitude to ##\epsilon##,

$$\vec E = -\nabla V = -vB\hat y$$

Now, this is where my issue is. We now have an electric field present in the resistor. Since we assumed the conductivity of the wires (and rod) is perfect, we know from Ohm's law that the electric field elsewhere in the circuit must be zero,

$$\vec E = \frac {\vec J} {\sigma} = \vec 0$$

However, since there is no time-changing magnetic field enclosed by the circuit, Faraday's law then reads,

$$\oint {\vec E \cdot d\vec l} = -\iint{\frac {\partial \vec B} {\partial t} \cdot d\vec S} = 0$$

This makes sense since we determined that the electric field from earlier in the resistor was conservative. However, given that this is a closed-loop line integral, that means there must be an electric field elsewhere in the circuit in order for the integral to vanish! That, or there is not in fact an electric field present in the resistor, which doesn't seem likely given that its conductivity is finite and it has a nonzero current density. Where is this other electric field, and why is it present despite Ohm's law from above showing that it's zero? Or, did I screw up somewhere (the more likely case)?

Thanks!

Source for image: http://labman.phys.utk.edu/phys222core/modules/m5/motional_emf.html

Consider the diagram attached to this post (source is linked at the bottom). Assume that all of the wires and the rod are perfect conductors, the magnetic field is uniform and time-independent (pointing into the page), and the velocity of the rod is constant (due to some applied force to the bar keeping it that way - we don't need to worry about that right now). We'll assume the self-inductance of the circuit is negligible. The positive x-direction is defined horizontally to the right in the figure, the positive y-direction is upwards, and the positive z-direction is out of the screen.

Now, as the rod moves to the right, the total emf in this circuit can be written as,

$$\epsilon = \oint ({\vec E + \vec v \times \vec B}) \cdot d \vec l$$

Now, given that the wires are thin and all of the charge stays within the wires, we may also write this as,

$$\epsilon = -\frac {d} {dt} \iint{\vec B \cdot d\vec S} = -\frac {d\Phi_B} {dt}$$

Now, my understanding is that this is

**not**Faraday's law. Rather, Faraday's law states,$$\oint {\vec E \cdot d\vec l} = -\iint{\frac {\partial \vec B} {\partial t} \cdot d\vec S}$$

As aforementioned, the magnetic field present is time-independent, so we should be able to conclude that,

$$\epsilon = \oint ({\vec v \times \vec B}) \cdot d \vec l$$

From here it is relatively straightforward to calculate the emf in the circuit. The leg of the circuit that is moving is the only part of the integral that contributes to the emf, and since the magnetic field and velocity are both uniform, we can easily see that,

$$\epsilon = vBd$$

Now, this is going to produce a current whose magnitude is given as,

$$I = \frac {vBd} {R}$$

This is all well and straightforward up to here. Now, this current produces an electric field in the resistor. We can see that this electric field in the resistor is conservative from the differential form of Faraday's law,

$$\nabla \times \vec E = -\frac {\partial \vec B} {\partial t} = \vec 0$$

Since the electric field is conservative, we can just determine it from the gradient of ##\epsilon##, with the potential difference across the resistor being equal in magnitude to ##\epsilon##,

$$\vec E = -\nabla V = -vB\hat y$$

Now, this is where my issue is. We now have an electric field present in the resistor. Since we assumed the conductivity of the wires (and rod) is perfect, we know from Ohm's law that the electric field elsewhere in the circuit must be zero,

$$\vec E = \frac {\vec J} {\sigma} = \vec 0$$

However, since there is no time-changing magnetic field enclosed by the circuit, Faraday's law then reads,

$$\oint {\vec E \cdot d\vec l} = -\iint{\frac {\partial \vec B} {\partial t} \cdot d\vec S} = 0$$

This makes sense since we determined that the electric field from earlier in the resistor was conservative. However, given that this is a closed-loop line integral, that means there must be an electric field elsewhere in the circuit in order for the integral to vanish! That, or there is not in fact an electric field present in the resistor, which doesn't seem likely given that its conductivity is finite and it has a nonzero current density. Where is this other electric field, and why is it present despite Ohm's law from above showing that it's zero? Or, did I screw up somewhere (the more likely case)?

Thanks!

Source for image: http://labman.phys.utk.edu/phys222core/modules/m5/motional_emf.html