Mutual Inductance between a coil and a long straight wire

[harsh22902]

Homework Statement

In the adjacent figure, the mutual inductance of the infinite straight wire and the
coil is M, while the self inductance of the coil is L. The current in infinite wire is
varying according to the relation I = at, where "a" is a constant and t is the time.
The time dependence of current in the coil is .

Relevant Equations

flux = L*i
flux due to a coupled coil = M* current through that coil

In the given question they have not provided the dimensions of the coil so I assumed it to be very close to the wire and having negligible dimensions compared to the wire . Then EMF induced in the coil due to the current in the wire comes out as M*a . Which when divided by resistance gives option a . But the correct answer is option D. Also the initial change in flux is due to current in the wire so I did not consider self inductance.

 

[Delta2]

You have to consider self inductance for this problem. There are two EMFs in this problem, one due to mutual inductance and one due to self inductance. We have to include both. The correct differential equation that describes this setup is $$M\frac{dI_1}{dt}+L\frac{dI_2}{dt}+I_2R=0$$ where $I_2$ is the current through resistor R, and we have included both EMFs (the EMF due to mutual inductance M $M\frac{dI_1}{dt}$ and the back EMF due to self inductance $L\frac{dI_2}{dt}$).
)The solution to the above differential equation is indeed none of the options provided.

 

[harsh22902]

The current I2 in the equation , is it due to emf induced by mutual inductance?

 

[Delta2]

The current I2 in the equation , is it due to emf induced by mutual inductance?

Well the mutual inductance is the primary cause if we can say that, but the self inductance also plays a role in its time dependence as that equation shows.

 

[harsh22902]

Also, the term I2*R that is actually the 'net' emf , right?

 

[Delta2]

Yes you can view it that way though it is actually the voltage drop in the resistance R but it is equal to the net emf as you say.

 

[harsh22902]

Understood it now, thank you !