# Can there be a potential difference perpendicular to induced electric field?

Here, since there is a time varying magnetic field, an induced electric field must be present which causes flow of current through the circuit. But OA is perpendicular to induced electric field so the potential difference between O and A must be zero but again current flows through the circuit so there must be some potential drop across the resistor. Please help me!

• There is an emf induced in the loop which produces an induced current in the loop which I assume is conducting. There is no evidence that the induced electric field is perpendicular to $OA$ which indeed is not the case as a current flows through the resistor. You have to be careful when using the concept of potential difference with regard to this example as the electric field is non-conservative. – Farcher Dec 15 '17 at 20:57
• Assuming that the magnetic field extends cylindrically with its centre at O, the induced electric fields form concentric circles with centre at O. Hence OA lies along the normal. – Sachin Dec 16 '17 at 4:34

I believe your question isn't so much about induced electric fields and Faraday's Law as it is about how electric fields work in circuits. I am going to try and explain the latter.

# summary:

As long as there is a net electromotive force $\mathcal{E}$ present somewhere in a circuit, the electric field on a conductor will even out to produce a uniform current. So even though the induced field $E_{induced}$ only contributes to the outside arc, the conductor sets up a uniform internal field $E_{internal}$ that extends to the rest of the circuit. It is this $E_{internal}$ that pushes electrons around. If you are a physics student, read Griffiths "Introduction to Electrodynamics" section 7.1.2 for a good summary.

# more details:

A conductor in an external electric field will set up its own internal electric field to cancel out the external one. See picture (A). If you can devise a scenario where only part of the conductor is in the external field, you can get current to flow continuously around the conductor according to Ohm's Law: $$\Sigma E = E_{external} - E_{internal} = \rho J$$ Where $\rho$ is the materials resistivity ($\rho=1/\sigma$, the conductivity) and $J$ the current density. In a perfect conductor $\rho$ is zero and so $E_{external} = E_{internal}$, but J is not necessarily zero. In practice the external electric field is produced any number of ways, for example inside a battery (picture (C)) or by induction.

# measuring the voltage:

Since voltage is the line integral of the electric field: $$V = \int E \cdot dl$$ if the wires are perfect conductors, $\Sigma E=0$ there. Inside the resistor, $\rho$ is not zero, and so you will find a non-zero Voltage. Faraday's Law will also give you the same Voltage.

• Nice explanation. I have one minor correction: your equation for Ohm's law should be $E = \rho J$, where $\rho$ is the resistivity of the material, not the conductivity. Making that change will make sense with the rest of your answer. – JM1 Dec 17 '17 at 17:14

There is an induced electric field in the loop producing a voltage $$V=\int \vec E d\vec r =-d\Phi/dt$$ across the outer circular sector. This causes a pile up of charges at the terminals of the resistor so that the induced loop voltage appears as a voltage $V$ across the resistor which drives the current.Thus an electric potential difference across the resistor is produced by induced electric charges.