# A long solenoid of length l, cross-sectional area A and having N1 turns (primary coil) has a small coil of N2

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A long solenoid of length l, cross-sectional area A and having N1 turns (primary coil) has a small coil of N2 turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.

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We assume the solenoid to be ideal and that all the flux from the solenoid passes through the outer coil C. For a steady current Is through the solenoid, the uniform magnetic field inside the solenoid is
B = µ0nIs ……………….. (1)
Then, the magnetic flux through each turn of the coil due to the current in the solenoid is
ΦCS = BA = (µ0nIs)(πR2) ……………… (2)
Thus, their mutual inductance is
M = $$\frac{N \Phi_{\mathrm{CS}}}{I_{\mathrm{S}}}$$ = µ0πR2nN ……………. (3)
Equation (2) is true as long as the magnetic field of the solenoid is entirely contained within the cross section of the coil C. Hence, M does not depend on the shape, size, or possible lack of close packing of the coil.
replacing n with N1/l and N with N2. M = µ0A = $$\frac{N_{1} N_{2}}{l}$$