I was reading Electromagnetic Induction chapter and a question came on my mind.. If I use a circuit with a battery and a single wound coil,and I pass 2A current through it,when the current will pass it...it will produce a back emf, and a induced current will be produced also,and the curent will try to oppose that cause of the change of magnetic flux. Here the cause of the magnetic flux is the 2A current, so the induced current will try to stop the current, Will it really stop the current instantly after the current passes through it?..or it will just reduce the current by a very small value?.If it is so, the motive of the induced current will be not satisfied right? As it will try to resist the change of magnetic flux,and just reducing the current by a small value will actually not satisfy it's motive fully So,the induced current will be as high as the actual current or much lower than that?

(I am using a DC battery as voltage source here)


1 Answer 1


The induced emf depends on the rate 0f change of current not the value of the current itself and it can never "win" in the sense of stopping the current completely because it id does then there is no change of current and hence no induced emf.

What an inductor does is to produced a induced emf which tries to slow down the rate at which the current is changing.

It is not clear from your question as to whether or not there is resistance in the circuit.
If there is resistance $R$ in the circuit then $\mathcal E_{\rm battery} + \mathcal E_{\rm induced} = IR \Rightarrow \mathcal E_{\rm battery} -L \dfrac {dI}{dt} = IR$ which is a first order differential equation with the solution $I = \frac{\mathcal E_{\rm battery}}{R} \left ( 1- \exp \left( -\frac {Rt}{L}\right)\right)$ assuming that the current in the circuit is zero when time $t$ is zero.
The current rises from zero at a slower and slower rate towards a current of $\frac{\mathcal E_{\rm battery}}{R}$.
You will not that the larger the inductance of the inductor $L$ the slower in the increase of current towards $\frac{\mathcal E_{\rm battery}}{R}$.


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