Doubt in electromagnetic induction

[Vatsal Goyal]

Homework Statement

suppose a coil is placed in a changing magnetic field and the circuit is not closed will the current induce in the coil

Homework Equations

The Attempt at a Solution

What I thought was that the current flows only in a closed circuit because it needs a potential difference (through a cell) which would require a closed path, but as there is no need of a cell to create potential difference in this case, the current should flow in the open circuit. But I checked the solution and it says that a potential difference will be induced but still no current would flow.

 

[Charles Link]

You did get an EMF from the changing magnetic field. If you don't have a closed loop, current will only flow very instantaneously, and in general, only a very minute amount. An electrostatic charge will build up from this minute current flow that creates a potential that balances the EMF. It only takes a very small amount of charge to get a considerable potential for wires and most other objects. Only in the case of a very large capacitor would the flow of current be very significant, e.g. (capacitor plates at opposite sides of the wire and connected to the wire), and even then it would be minimal except with an extremely large capacitor. In general, you need a closed loop to get significant currents.

 

[Vatsal Goyal]

Thank you for your answer and sorry for replying late.

I didn't understand this statement of yours completely, can you please elaborate?

It only takes a very small amount of charge to get a considerable potential for wires and most other objects

Also would this current be significant enough to be detected by a galvanometer?

If you don't have a closed loop, current will only flow very instantaneously, and in general, only a very minute amount.

 

[Charles Link]

Approximate calculations can be done using voltages, and if you take a small conductive sphere, $V=\frac{Q}{4 \pi \epsilon_o r}$ . If $r=1.0$ E-3 m (1 mm), you can see for a very small $Q$ you can get a large $V$. $\\$ The actual equilibrium in the conductor occurs when the electric field is zero everywhere, i.e. when the electrostatic electric field is equal and opposite the induced electric field, whose integral is the EMF. Basically you can balance the EMF with the potential of the electrostatic part of the electric field, (the voltage (potential) is the integral of the electrostatic field), for approximate results. It is because so little charge is necessary to establish these voltages that our households don't use any significant electricity when nothing is plugged into the outlets. $\\$ There may be galvanometers that can measure small amounts of charges like you are referring to, but a standard volt-ohm meter could not measure them.

 

[Vatsal Goyal]

Approximate calculations can be done using voltages, and if you take a small conductive sphere, $V=\frac{Q}{4 \pi \epsilon_o r}$ . If $r=1.0$ E-3 m (1 mm), you can see for a very small $Q$ you can get a large $V$. $\\$ The actual equilibrium in the conductor occurs when the electric field is zero everywhere, i.e. when the electrostatic electric field is equal and opposite the induced electric field, whose integral is the EMF. Basically you can balance the EMF with the potential of the electrostatic part of the electric field, (the voltage (potential) is the integral of the electrostatic field), for approximate results. It is because so little charge is necessary to establish these voltages that our households don't use any significant electricity when nothing is plugged into the outlets. $\\$ There may be galvanometers that can measure small amounts of charges like you are referring to, but a standard volt-ohm meter could not measure them.

Thank you, I think I got it!