This is what i've drawn...assume that switch is working rapidly.the first figure shows the direction of magnetic field due to current from battery. Which is proper.but i get confused about second figure below it. Is it the correct polarity of self induced emf ? What will be the direction of magnetic field due to self induced emf...i am stuck..please explain me clear concept about it.
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\$\begingroup\$ Put a 100 kΩ resistor in parallel with the inductor so the current can go somewhere. And remember that an inductor (or rather its magnetic field) resists changes with current. - What you will end up with is essentially a current source and a resistor. Which side will be very positive / negative? \$\endgroup\$– Harry SvenssonCommented Nov 30, 2017 at 17:08
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\$\begingroup\$ Thank you so,Is it because of self induced emf that inductor resist the change? \$\endgroup\$– user133933Commented Nov 30, 2017 at 17:12
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1 Answer
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Your indication of the voltage polarity after the switch opens is correct, however your understanding of the magnetic field at that point is not.
The magnetic field existed before the switch opened and at that moment remains the same, meanwhile, the current still wants to flow through the inductor.
The magnetic field it will decay very quickly as the current decays in inductor.
simulate this circuit – Schematic created using CircuitLab
Until it does, and as you can see from the image above, the only current path for the inductor to cycle that current, is through the parasitic impedance of the air around it. The voltage drop across that impedance drops the left side of the inductor to a negative value compared to the right side.
This is why, when switching inductors, we use fly-back diodes. The voltage on the left side of the diode in the image below will drop to the point where the diode turns on and allows the current in the inductor, and the associated magnetic field, to safely decay.
