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CBSE Class 12 Physics 2014 Outside Delhi Set 1 Paper

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Question : 20 of 30
Marks: +1, -0
Define the term 'mutual inductance' between the two coils.
Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length ll and radii r1r_1 and r2(r2≫r1)r_2 (r_2 \gg r_1). Total number of turns in the two solenoids are (N1.(N_1. and N2)N_2) respectively.
Solution:  
Mutual inductance of two coils is equal to the e.m.f. induced in one coil when rate of change of current through the other coil is unity.
Mutual inductance of two co-axial solenoids : Consider two long co-axial solenoid each of length 1 with number of turns N1N_1 and N2N_2 wound one over the other. Number of turns per unit length in solenoid, n=N1ln=\frac{N_1}{l}. If I1I_1 is the current flowing in primary solenoid, the magnetic field produced within this solenoi
B1=μ0N1I1lB_1=\frac{\mu_0 N_1 I_1}{l}
The flux linked with each turn of inner solenoid coil is
ϕ2=N2ϕ\phi_2=N_2\phi
=N2B1A2=N_2 B_1 A_2
=N2(μ0N1I1l)A2=N_2 \left(\frac{\mu_0 N_1 I_1}{l}\right) A_2
Mutual Inductance,
M21=(μ0N1N2l)A2M_{21}= \left(\frac{\mu_0 N_1 N_2}{l}\right) A_2
If n1n_1 is number of turns per unit length of outer solenoid and r2r_2 is radius of inner solenoid, then
M=μ0n1N2πr22M=\mu_0 n_1 N_2 \pi r_2^2
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