Question

Imagine that we have two solenoids. Solenoid 1 has Nį loops, a length li, and a radius ai, while solenoid 2 has N2 loops, a length l2, and a radius a2. Assume that l2 <lı and a2 > a1. The solenoids are arranged as shown (here we just show the cross-section in the page, but both consist of full, circular loops). For parts a-d, treat the solenoids as ideal. a. [4 points] If a current 12 flows in solenoid 2, what is the total flux through the loops of solenoid 1, 01? b. [2 points] Calculate the mutual inductance, M12, for the emf in loop 2 solenoid 1 due to a change of the current in the solenoid 2 from part loop 1 J! OU lol 10 ! a. li VOO000000 Ⓡ! J STOTTA c. [4 points] Now suppose that a current 11 flows in solenoid 1. What is the total flux through the loops of solenoid 2, 02? d. [2 points] Calculate the mutual inductance, M21, for the emf in solenoid 2 due to a change of the current in solenoid 1 from part c. e. [3 points] Now imagine that solenoid 2 is not ideal. How would your answers to parts b and d change and why?. a2 ai

Answer 1

Solution:

Part (a): If a current I₂ in flowing in the solenoid 2, then a constant magnetic field produced inside the solenoid 2 and given by

(1)

B2 = MON21

So, the total flux passing through soleniod 1 is given by

$\phi _{1}=flux \ through \ one \ loop \times \ No. \ of \ total \ loops\\\\ \Rightarrow \phi _{1}=B_{2}A_{1}\cos 0\times N_{1}\\\\ \Rightarrow \phi _{1}=\mu _{0}N_{2}I_{2}\times \pi a_{1}^{2}\times N_{1}\\\\ \Rightarrow \phi _{1}=\pi \mu _{0}N_{1}N_{2}I_{2}a_{1}^{2}\\\\$ (2)

Part (b): Emf induced in the solenoid 1 is proportional to the rate of change of current in the solenoid 2 w.r.t. time and given by

$e_{1}\propto \frac{dI_{2}}{dt}\\\\ \Rightarrow e_{1}=-M_{12}\frac{dI_{2}}{dt}$    (3)

Also,

$e_{1}=-\frac{d\phi _{1}}{dt}$ (4)

Putting eqⁿ (4) in eqⁿ (3), we find

$M_{12}=\frac{d\phi _{1}}{dI_{2}}\\\\$ (5)

Ucing eqⁿ (2) in eqⁿ (5), we find the mutual inductance in solenoid 1 due to the current change in the solenoid 2 as

$M_{12}=\pi \mu _{0}N_{1}N_{2}a_{1}^{2}$ (6)

Part (c): Similarly, If a current I₁ in flowing in the solenoid 1, then a constant magnetic field produced inside the solenoid 1 and given by

$B_{1}=\mu _{0}N_{1}I_{1}$ (7)     

So, the total flux passing through soleniod 2 is given by

$\phi _{2}=flux \ through \ one \ loop \times \ No. \ of \ total \ loops\\\\ \Rightarrow \phi _{2}=\int Bds\cos 0\times N_{2}\\\\ \Rightarrow \phi _{2}=N_{2}\left [\int_{0}^{A_{1}}B_{1}ds+\int_{A_{1}}^{A_{2}}0ds \right ]\\\\ \left ( \because \ magnetic \ field \ outside \ the \ solenoid 1 \ is \ zero \ \right )\\\\ \Rightarrow \phi _{2}=N_{2}\left [B_{1}A_{1}+0 \right ]\\\\ \Rightarrow \phi _{2}=N_{2\times }\mu _{0}N_{1}I_{1}\times \pi a_{1}^{2}\\\\ \Rightarrow \phi _{2}=\pi \mu _{0}N_{1}N_{2}I_{1}a_{1}^{2}\\\\$ (8)

Part (d): Emf induced in the solenoid 2 is proportional to the rate of change of current in the solenoid 1 w.r.t. time and given by

$e_{2}\propto \frac{dI_{1}}{dt}\\\\ \Rightarrow e_{2}=-M_{21}\frac{dI_{1}}{dt}$ (9)

Also,

$e_{2}=-\frac{d\phi _{2}}{dt}$ (10)

Putting eqⁿ (10) in eqⁿ (9), we find

$M_{21}=\frac{d\phi _{2}}{dI_{1}}\\\\$ (11)

Ucing eqⁿ (8) in eqⁿ (11), we find the mutual inductance in solenoid 2 due to the current change in the solenoid 1 as

$M_{21}=\pi \mu _{0}N_{1}N_{2}a_{1}^{2}$    (12)