Considering a moving coil galvanometer that isn't connected in a circuit, say I deflect it and release it. The coil should start oscillating. But now, what if I suddenly connect the two ends of the coil of the galvanometer? Is it correct to say that an emf is induced (in accordance to Lenz's law), and a current is produced such that a magnetic field is produced, that leads to no torque acting on the coil, and the oscillation stops? Not very sure about this. What would exactly happen?


[…] say I deflect it and release it. The coil should start oscillating.

Regular current meters are damped in order to get a sensible readout, I am not sure how much damping would be in such a galvanometer, probably a lot less. So it might oscillate freely for a while.

But now, what if I suddenly connect the two ends of the coil of the galvanometer.

From energy conservation one can estimate that the oscillation will die out quickly because the wire has a low resistance and will therefore burn a lot of power, quickly converting the kinetic/potential energy of the coil into heat.

Is it correct to say that an emf is induced (in accordance to Lenz's law) […]

The EMF is always there, even without the wire connecting the ends. It is just a voltage. There was no current before because there was no wire.

[…] and a current is produced such that a magnetic field is produced […]

Yes. The wire allows the already existing voltage to drive a current now.

[…] such that a magnetic field is produced […]

And that magnetic field will be opposing the external field that the coil is embedded in.

[…] that leads to no torque acting on the coil, and the oscillation stops?

If there was no torque, the oscillation would continue. The stopping of the oscillation is done by torque that is opposing the current direction of rotation.

I'd say that the opposing magnetic field that is created by allowing the current to flow will create a torque which opposes the rotation and therefore lets the oscillation die out.

  • $\begingroup$ Why would the magnetic field produced oppose the existing field? If I deflect the coil, the area of the coil through which the existing field is passing would decrease, correct? Hence decreasing the flux through the coil. According to Lenz's law, wouldn't an emf be induced in such a way that the flux increases (opposing the change)? $\endgroup$ – Shreya Aug 20 '17 at 6:12

A conventional moving coil galvanometer is really equivalent to a spring-mass system with rotational motion.
So it is the moment of inertia of the coil and former on which the coil is mounted and the springs which provide a restoring couple which are important.
As with all such systems there is damping and in a moving coil galvanometer the damping is arranged to be just less than critical damping so that the pointer on the galvanometer will reach the "reading" position in the smallest amount of time.
This slightly less than critical amount of damping is produced by careful design of the galvanometer and can be split into two parts.
Frictional damping due to air resistance, friction at the bearings etc and electromagnetic damping the coil being wound on an aluminium former.
As the former moves in the magnetic field of the galvanometer's permanent magnet an emf is induced in the aluminium (Faraday) and as aluminium is a conductor there is an induced (eddy) current in the former which tries to oppose the motion producing it (Lenz).
So the rotational kinetic energy of the coil/former system is converted into heat in the aluminium former.

Now by adding a resistor across the terminals of a galvanometer you have introduced an extra form of electromagnetic damping.
The coil moving in the magnetic field of the galvanometer's permanent magnet will induce an emf and if there is a complete electrical circuit formed by connecting the terminals of the galvanometer a current will flow which tries to oppose the motion producing it.
Again rotational kinetic energy is converted into heat and some more electromagnetic damping is introduced and this time the coil/former-sring system is likely to be over damped resulting in a "sluggish" movement of the pointer.

Demonstrations of Faraday's law using a moving magnet and a coil connected to a galvanometer can often be improved by introducing a series resistor in the coil and galvanometer circuit as this will reduce the electromagnetic damping of the galvanometer and potentially increase the deflection which is achieved on the galvanometer.

A ballistic galvanometer is designed with the minimum of damping in mind and as such its coil is wound on a non conducting former so there is no electromagnetic damping due to eddy currents in the former.
To reduce the effects of any mechanical damping which is present the coil is suspended from a torsion wire and has a large moment of inertia.
Such a galvanometer is under damped and when it is used care has to be taken to ensure that the resistance of any external circuit connected to is is high enough to reduce to effect of electromagnetic damping.
So a search coil used to measure magnetic fields would normally have a seires resistor connected to it.

A ballistic galvanometer is a device which can be used to demonstrate the three regimes of damping (under, critical and over) simply by varying the resistance placed across the terminals.
There are many reference to such experiments/demonstrations on the Internet and in textbooks and here is one.


Such a galvanometer is called a 'ballistic galvanometer' and would typically be read while moving... and with the galvanometer wires disconnected (open).

If the terminals of the galvanometer are shorted, then the resistance of the galvanometer wiring is excited by the EMF that you describe, and the resulting current does generate torque that brings the galvanometer to a halt (exponentially, so only friction or Brownian motion cause a true zero of the angular velocity). In the absence of a closed circuit, the galvanometer motion is only slowly damped by air viscosity, imperfect elasticity of the springs, and pivot friction.

A quartz-fiber suspension (no pivot, the fiber acts as support and spring) ballistic galvanometer can take many minutes to stop swinging.

  • $\begingroup$ How does the resulting current generate a torque? $\endgroup$ – Shreya Aug 20 '17 at 6:30
  • $\begingroup$ A galvanometer contains a fixed magnet, and a coil of wire. Current in the coil makes an electromagnet, which (like a compass needle) has torque because of the fixed magnet next to it. $\endgroup$ – Whit3rd Aug 20 '17 at 7:08
  • $\begingroup$ It isn't necessary that that torque is equal to the restoring torque caused by the spring, right? Also, is it correct to say then that the coil will stop moving completely when the north of the permanent magnet (of the galvanometer) is aligned with the south of the electromagnet (coil)? $\endgroup$ – Shreya Aug 20 '17 at 7:49
  • $\begingroup$ When the galvanometer stops moving, it HAS no 'south of the electromagnet' that can be identified. Magnetic torque on the coil, like the motion of the coil, is oscillatory (and always decelerates the coil). $\endgroup$ – Whit3rd Aug 20 '17 at 8:00
  • $\begingroup$ Okay... so the coil never stops completely? $\endgroup$ – Shreya Aug 20 '17 at 12:27

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