Name: Electric Field: Time Varying Magnetic Field
Uploaded: 2019-07-19
Duration: 8 min 32 s
Description: Did you know that the cause of the movement of charges in a stationary circular conducting coil placed in a time-varying magnetic field is due to an induced...

Question

Two long cylindrical coils with uniform windings of the same length and nearly the same radius have inductances $L_{1}$ and $L_{2}$ . The coils are coaxially inserted into each other and connected to a current source as shown in. The directions of the current in the circuit and in the turns are shown by arrows.

Determine the inductance $L$ of such a composite coil.

Question Image

Determine the inductance $L$ of such a composite coil.

Accepted Answer

Let $N_{1}$ be the number of turns of the coil of inductance $L_{1}$ , and $N_{2}$ be the number of turns of the coil of inductance $L_{2}$ . It should be noted that the required composite coil of inductance $L$ can be treated as a coil with $N = N_{1} + N_{2}$ turns. If the relation between the inductance and the number of turns is known, $L$ can be expressed in terms of $L_{1}$ and $L_{2}$ . For a given geometrical configuration of the coil, such a relation must actually exist because

inductance is determined only by geometrical configuration and the number of turns of the coil (we speak of long cylindrical coils with uniform winding). Let us derive this relation.

From the superposition principle for a magnetic field, it follows that the magnetic field produced by a current $I$ in a coil of a given size is proportional to the number of turns in it. Indeed, the doubling of the number of turns in the coil can be treated as a replacement of each turn by two new closely located turns. These two turns will produce twice as strong a field as that produced by a single turn since the fields produced by two turns are added. Therefore, the field in a coil with twice as many turns is twice as strong. Thus, $B \propto N$ ( $B$ is the magnetic induction, and the current is fixed). It should be noted that the magnetic flux embraced by the turns of the coil is

$Φ = B N S \propto B N \propto N^{2} .$

It remains for us to consider that

$L = \frac{Φ}{I} \propto N^{2}$

Thus, we obtain $L = k N^{2}$ for a given geometry. Further, we take into account that $N_{1} = \sqrt{\frac{L_{1}}{k}}, N_{2} = \sqrt{\frac{L_{2}}{k}}$ and hence $L = k {(N_{1} + N_{2})}^{2}$ . Consequently,

$L = k {[\sqrt{\frac{L_{1}}{k}} + \sqrt{\frac{L_{2}}{k}}]}^{2}$

$L = k [\frac{L_{1}}{k} + \frac{L_{2}}{k} + 2 \frac{\sqrt{L_{1} L_{2}}}{K}]$ .

$L = L_{1} + L_{2} + 2 \sqrt{L_{1} L_{2}} .$

S S Krotov Solutions for Chapter: Electricity and Magnetism, Exercise 1: Exercise

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