# Faraday's law with a battery

So I've recently learned about Faraday's law of induction and I'm still confused about a particular situation. Suppose there is an electric circuit: a circular loop of radius $$r$$ composing of a battery and a resistance. The battery generates an emf $$\mathcal{E}$$ and due to Faraday's law of induction $$\mathcal{E} = -\frac{d\Phi_B}{dt}$$ with $$\Phi_B = \oint B \cdot dS = B \cdot \pi r^2$$ then, by solving the equation we have $$\Phi_B = -\mathcal{E}t + C$$ which gives $$B = \frac{-\mathcal{E}t + C}{\pi r^2}$$ which doesn't make sense. This result implies that the magnetic field is generated indefinitely (i.e. $$B$$ increases as I wait longer, depending on $$t$$). Therefore, I'd like to ask for an explanation on what I misunderstood. Thank you in advance.

• The emf of the battery is not an emf induced by a changing magnetic field. Faraday's law does not apply in this case.
– d_b
Commented Apr 10, 2022 at 20:26
• @d_b If it does not work, how can we describe the resulting magnetic field due to a battery? i.e. how much can a loop of current generate a magnetic field, and when it's generated, is the emf going to be induced backward on the circuit? Commented Apr 10, 2022 at 20:34
• When the circuit is first connected, the change in current leads to a changing magnetic flux, so there will be an emf induced that counters the emf of the battery. Eventually the circuit will reach a steady state where the emf of the battery drives a steady current and there is no changing flux or induced emf.
– d_b
Commented Apr 10, 2022 at 20:51

Assuming a series circuit and no initial current.

The circuit is completed anly at that instant is the current zero and the induced emf equal in magnitude to the emf of the battery (your first equation).

Subsequent to that there is a potential difference across the resistance in the circuit which you have to include in any circuit calculation.

As the current increases the voltage across the resistor increases and the induced emf decreases as the rate of increase of current (and hence the rate of change of magnetic flux) in the circuit decreases.

• So you are saying that at the equilibrium (i.e. when everything is done), the voltage across the resistor is at the full voltage, the rate of change of current is now zero, the induced emf is now zero, and so the magnetic flux doesn't change anymore? I think I still don't get the notion of induced emf. If a battery is connected with voltage V, does that mean that the emf inside the circuit is always V? Commented Apr 10, 2022 at 20:47
• And it is that $V$ which drives the current through the resistance. Commented Apr 11, 2022 at 2:12

Note that you are solving the integral for the flux of the magnetic field over a closed surface and not over a surface that does not close on itself. They are distinguished by the "o" in the integral symbol ($$\oint$$ and$$\int)$$. You interchanged two formulas: $$\Phi_B = \int_C {\mathbf{B} \cdot \mathrm d\mathbf{A} } \\ \oint_S {\mathbf{B} \cdot \mathrm d\mathbf{A}} = 0$$

A more correct way would be: $$\varepsilon =-\frac{\mathrm d\Phi_B}{\mathrm dt} \rightarrow \varepsilon= -\frac{\mathrm d}{\mathrm dt}\int_C {\mathbf{B} \cdot \mathrm d\mathbf{A} } = \int_C {-\frac{\mathrm d \mathbf{B}}{\mathrm dt} \cdot \mathrm d\mathbf{A} }$$

You would have to know how the $$\mathbf B$$ field and $$\varepsilon$$ changes with time for you to be able to integrate.

Just for fun. You could derive one of the four Maxwell's Equation by using this formula: $$\varepsilon = \int{\mathbf E \cdot \mathrm d \mathbf r} \\ \int_C{\mathbf E \cdot \mathrm d \mathbf r} = \int_C {-\frac{\mathrm d \mathbf{B}}{\mathrm dt} \cdot \mathrm d\mathbf{A} } \\ \int_C{\nabla \times\mathbf E \cdot \mathrm d \mathbf A} = \int_C {-\frac{\mathrm d \mathbf{B}}{\mathrm dt} \cdot \mathrm d\mathbf{A} } \\ \nabla \times\mathbf E = -\frac{\mathrm d \mathbf{B}}{\mathrm dt}$$

• Oh, I noticed my mixed-up of the symbol, right. Thank you for the clarification. Commented Apr 10, 2022 at 20:43