Answer:
[tex]E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)[/tex]
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:
[tex]\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I[/tex]
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:
[tex]\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)[/tex]
Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)
Using:
[tex]B_{1}(t) =\mu _{o} nI(t)\\[/tex] --- (2)
Flux generated due to magnetic field B₁
[tex]\phi _{1}(t) = \oint \overrightarrow {B_{1}}.dA\\[/tex] ---(3)
area of solenoid = [tex]A = \pi \rho^{2}[/tex]
substituting (2) in (3)
[tex]\phi _{1}(t) = \mu_{o} \pi \rho^{2} n_{1}I_{1}(t)[/tex] ----(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.
[tex]E_{2} (t) = -n_{2}L\frac{d \phi_{1}}{dt}[/tex] ---- (5)
substituting (4) in (5)
[tex]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)[/tex]
