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Question
Consider two long co-axial solenoids S1 and S2, each of length I (>> r2) and of radius r1 and r2(r2 > r1). The number of turns per unit length are n1 and n2 respectively. Derive an expression for mutual inductance M12 of solenoid S1 with respect to solenoid S2. Show that M21 = M12.
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Solution
Let l = length of each solenoid
r1, r2 = radii of the two solenoids
A = `pi r_1^2`
A = area of cross-section of inner solenoid S1
N1, N2 = number of turns in the two solenoids
First we pass a time varying current I2 through S2.
The magnetic field set up inside S2 due to I2 is:
B2 = μ0n2I2
Where n2 = N2/l = the number of turns per unit length of S2
Total magnetic flux linked with the inner solenoid S1 is:
Φ1 = B2AN1 = μ0n2I2AN1
∴ Mutual inductance of coil 1 with respect to coil 2 is;
M12 = `phi_1/l_2`
= μ0n2AN1
= `(mu_0N_1N_2A)/l`
We now consider the flux connected to the outer solenoid S2 as a result of the inner solenoid’s current I1. S1. The field B1 due to I1 is constant inside S1 but zero in the annular region between the two solenoids.
Hence B1 = μ0n1I1
Where n1 = N1/l = the number of turns per unit length of S1.
Total flux linked with the outer solenoid S2 is:
Φ2 = B1AN2 = μ0n1I1AN2
= `(mu_0N_1N_2AI_1)/l`
∴ Mutual inductance of coil 2 with respect to coil 1 is:
M21 = `(mu_0N_1N_2A)/l`
Clearly, M12 = M21 = M
