Question

Now find the electromotive force E2(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1. Remember that both solenoids have length L.

Answer 1

Answer:

$E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)$

Explanation:

Consider two solenoids that are wound on a common cylinder as shown in fig. 1. Let the cylinder have radius 'ρ' and length 'L' .

No. of turns of solenoid 1 = n₁

No. of turns of solenoid 1 = n₂

Assume that length of  solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length:

$\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I$

We will consider the field that arises from solenoid 1, having n₁  turns per unit length. The magnetic field due to solenoid 1 passes through solenoid 2, which has n₂ turns per unit length.

Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction:

$\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)$

Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)

Using:

                                  $B_{1}(t) =\mu _{o} nI(t)$ --- (2)

                           

Flux generated due to magnetic field B₁

                      $\phi _{1}(t) = \oint \overrightarrow {B_{1}}.dA$ ---(3)

area of solenoid = $A = \pi \rho^{2}$

substituting (2) in (3)

                       $\phi _{1}(t) = \mu_{o} \pi \rho^{2} n_{1}I_{1}(t)$ ----(4)

We have to find electromotive force E₂(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1, i.e.

                       $E_{2} (t) = -n_{2}L\frac{d \phi_{1}}{dt}$ ---- (5)

substituting (4) in (5)

$E_{2} (t) = -n_{2}L\frac{d }{dt}(\mu_o} \pi \rho^{2} n_{1}I_{1}(t))\\E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)$