Mutual inductance between two long co-axial solenoids:
Consider two long co-axial solenoids of same length 1. The length of these solenoids is large when compared to their radii so that the magnetic field produced inside the solenoids is uniform and the fringing effect at the ends may be ignored. Let A1 and A2 be the area of cross section of the solenoids with A1 being greater than A2 . The turn density of these solenoids are n1 and n2 respectively.
Mutual inductance of two long co-axial selonoids
Let i1 be the current flowing through solenoid 1, then the magnetic field produced inside it is B1 = μ0n1i1.
As the field lines of \(\vec B\) are passing through the area bounded by solenoid 2, the magnetic flux is linked with each turn of solenoid 2 due to solenoid 1 and is given by
since θ = 0°
The flux linkage of solenoid 2 with total turns N2 is
N2 Φ21 = (n2 l)(μ0n1i1 )
since N2 = n2l
N2 Φ21 = (μ0n1n1 A2l)i1 ….. (1)
From equation of mutual induction
N2Φ21 = M21i1 …… (2)
Comparing the equations (1) and (2),
M21 = μ0n1n1 A2l ….. (3)
This gives the expression for mutual inductance
M21 of the solenoid 2 with respect to solenoid
1. Similarly, we can find mutual inductance M21 of solenoid 1 with respect to solenoid 2 as given below.
The magnetic field produced by the solenoid 2 when carrying a current i2 is
B = μ0n2i2
This magnetic field B2 is uniform inside the solenoid 2 but outside the solenoid 2, it is almost zero. Therefore for solenoid 1, the area A2 is the effective area over which the magnetic field B2 is present; not area A2 Then the magnetic flux Φ12 linked with each turn of solenoid 1 due to solenoid 2 is
The flux linkage of solenoid 1 with total turns N1 is
[Since N1 = n1l]
Therefore, we get
∴ M12 = μ0n1n2A2l ……. (4)
From equation (3) and (4), we can write
M12 =M21=M ……. (5)
In general, the mutual inductance between two long co-axial solenoids is given by
M= μ0n1n2 A2l ……. (6)
If a dielectric medium of relative permeability’ pr is present inside the solenoids, then
M = μn1n2A2l
or M = μ0μr n1n2 A2l