# Problem with Faraday's Law for a Circuit with Battery

From Maxwell's equations, we know, $$\nabla \times \vec E=-\frac{\partial \vec B}{\partial t}\Rightarrow\oint \vec E\cdotp d\vec l=-\int\frac{\partial \vec B}{\partial t}\cdotp d\vec a \tag{1}\label{1}$$ and from the definition of EMF, $$\varepsilon=\oint \vec f\cdotp d\vec l \tag{2}\label{2}$$ where, $$\vec f$$ is the force per unit charge. Now, consider a simple circuit with just a battery and a resistance. The circuit is placed in a time varying magnetic field. So, $$\vec f=\vec f_s+\vec E$$, where $$\vec f_s$$ is the force per unit charge due to the battery. So, the EMF $$\varepsilon=\oint(\vec f_s+\vec E)\cdotp d\vec l=\oint\vec f_s\cdotp d\vec l+\oint\vec E\cdotp d\vec l=\oint\vec f_s\cdotp d\vec l-\int\frac{\partial \vec B}{\partial t}\cdotp d\vec a \tag{3}\label{3}$$ But from Faraday's law, we know that the EMF, $$\varepsilon=-\frac{d\phi}{dt}=-\int\frac{\partial \vec B}{\partial t}\cdotp d\vec a \tag{4}\label{4}$$ Combining these two expressions of $$\varepsilon$$, we get $$\oint\vec f_s\cdotp d\vec l-\int\frac{\partial \vec B}{\partial t}\cdotp d\vec a=-\int\frac{\partial \vec B}{\partial t}\cdotp d\vec a \tag{5}\label{5}$$ $$\Rightarrow \oint\vec f_s\cdotp d\vec l=0$$ But, $$\oint\vec f_s\cdotp d\vec l$$ is actually the voltage across the battery terminals which shouldn't be zero. Where is the problem with this analysis?

• how did you lose $\oint \vec E\cdotp d\vec l$? Commented Feb 11, 2023 at 20:33
• I used Maxwell's equation (the very first equation here) to lose it. Commented Feb 11, 2023 at 20:40
• I have been asking myself the same question as ＠Mobin_Haque for a long time. Electro-chemistry says that the Gibbs energy is the origin of the electromotive force of batteries, but I am not convinced. I think it is not yet possible to derive the electromotive force of a battery from the fundamental laws of physics. Commented Feb 11, 2023 at 22:21

You're confusing induced EMF with total EMF, which is sum of all contributions to EMF, including the induced EMF(due to induced electric field, described by the Faraday law) and the battery EMF (due to electrochemical processes in the battery, not described by the Faraday law).

Net EMF due to both induced field and battery can be written as

$$\mathscr{E}_{net} = \mathscr{E}_{induced} + \mathscr{E}_{battery}$$ $$\mathscr{E}_{net} = \oint \mathbf E_{induced} \cdot d\mathbf s + \mathscr{E}_{battery}.$$ $$\mathscr{E}_{net} = -\frac{d}{dt}\int \mathbf B \cdot d\mathbf S + \mathscr{E}_{battery}.$$

If resistance of the circuit is $$R$$, then Kirchhoff's second circuital law predicts that current due to both sources of EMF will be

$$I = \frac{\mathscr{E}_{net}}{R}.$$

• Thank you for your clarification. No wonder the voltage across the battery terminals was found to be zero in my analysis !! Commented Feb 12, 2023 at 10:08

The Equation (4), third equation up from the bottom: Faraday's law gives the emf due to changing magnetic flux. You seem to be equating $$-\frac{d\Phi}{dt}$$ to the total emf in the circuit, including that of the battery.

• Thank you for your answer. I understand my mistake now. Commented Feb 12, 2023 at 10:07