Question

Suppose that we have two solenoids of the same length. Their diameters differ only to the extent to which one can be fitted onto the other. The inductances of the two solenoids can be considered the same and equal to $L$ . Here are the ways in which the solenoids can be connected:

( $1$ ) the solenoids are connected in series and are separated by a large distance;

( $2$ ) the solenoids are connected in parallel and are separated by a large distance;

( $3$ ) the solenoids are connected in series, one is fitted onto the other, and the senses of the turns coincide;

( $4$ ) the solenoids are connected in parallel, one is fitted onto the other, and the senses of the turns coincide;

( $5$ ) the solenoids are connected in series, one is fitted onto the other, and the senses of the turns are opposite;

( $6$ ) the solenoids are connected in parallel, one is fitted onto the other, and the senses of the turns are opposite.

Determine the total inductance for each of the above cases.

Accepted Answer

$(1)$ The system can be considered as being a new solenoid whose length is twice as large as that of one solenoid, with a density of turns the same as that in one solenoid and with the same cross-sectional area. Since the inductance of one solenoid is $L = μ_{0} N_{0} V$ , where $N_{0}$ is the number of turns per unit length, and $V$ is the solenoid volume, which in this case is twice the volume of one solenoid, we have $L_{1} = 2 L_{0}$ . The same result can be achieved by considering the self-induction emf that is generated in the two solenoids connected in series. The total emf is equal to the sum of the emf's generated in each solenoid; hence,

$E_{1} = - 2 L_{0} \frac{d I}{d t}$ ,

which yields $L_{1} = 2 L_{0}$ .

$(2)$ When the solenoids are connected in parallel, the self-induction emf in each solenoid is

$E_{1} = - L_{0} \frac{d (\frac{I}{2})}{d t} = - \frac{1}{2} L_{0} \frac{d I}{d t}$ .

Because the solenoids are connected in parallel, the total emf has the same value. Thus, with a current $L$ in a circuit that is external with respect to the solenoid, the induced emf is one-half the value for the inductance $L_{0}$ . Hence,

$L_{2} = \frac{1}{2} L_{0}$ .

$(3)$ In this case, the number of turns per unit length is twice as large as that of one solenoid, and since the inductance is proportional to $6$ , we have (provided that the current remains unchanged)

$L_{3} = 4 L_{0}$ .

$(4)$ If one solenoid is fitted onto the other and the senses of the turns coincide and the solenoids are connected in parallel, the current through each solenoid is $\frac{I}{2}$ if the current in the circuit is $I$ , while the flux associated with current $\frac{I}{2}$ is $\frac{ϕ}{2}$ . The total flux is $ϕ$ and the flux coupled with each solenoid is $ϕ' = ϕ N_{0}$ . In each solenoid, there is generated a self-induction emf equal to the one induced in a separate solenoid when current, $I$ varies. Since the solenoids are connected in parallel, this emf is the common emf of both solenoids. Hence, $L_{4} = L_{0}$ .

$(5, 6)$ In both cases, the induction flux in the solenoids is zero, so that,

$L_{5} = L_{6} = 0$ .

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