Faraday's law with a battery

Question

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.

Answer 1

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.

Answer 2

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: $$ \varepsilon = \int{\mathbf E \cdot \mathrm d \mathbf r} \\ \int_C{\mathbf E \cdot \mathrm d \mathbf r} = \int_C {-\frac{\mathrm d \mathbf{B}}{\mathrm dt} \cdot \mathrm d\mathbf{A} } \\ \int_C{\nabla \times\mathbf E \cdot \mathrm d \mathbf A} = \int_C {-\frac{\mathrm d \mathbf{B}}{\mathrm dt} \cdot \mathrm d\mathbf{A} } \\ \nabla \times\mathbf E = -\frac{\mathrm d \mathbf{B}}{\mathrm dt} $$