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Two concentric circular coils, one of small radius r_1 and the other of large radius r_2, such that r_1 lt lt r_2, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. |
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Answer» Solution :Consider two EXTREMELY long co-axial CIRCULAR coils (or solenoids) of equal length L, with smaller one inside the bigger one. Let small and big coils be coil no. 1 and coil no. 2 respectively. Their common area of cross section is suppose `a=pir_1^2`. When `I_1` amount of current is passed through `N_1` turns of small coil, magnetic FLUX linked with big coil is, `Phi_2=N_2B_1 a cos 0^@` (Here, magnetic field is present only inside the small coil and so cross-sectional area of small coil is considered). `therefore Phi_2=N_2((mu_0N_1I_1)/l)a` `therefore Phi_2/I_1=(mu_0N_1N_2a)/l` Here `Phi_2/I_1=M_21`=mutual inductance of second coil towards FIRST coil. `therefore M_21=(mu_0N_1N_2a)/l`....(1) Similarly when `I_2` amount of current is passed through `N_2` no. of turns of big coil, magnetic flux linked with small coil is, `Phi_1=N_1B_2a` `=N_1((mu_0N_2I_2)/l)a` `therefore Phi_1/I_2=(mu_0N_1N_2a)/l` Here `Phi_1/I_2=M_12`=mutual inductance of first coil towards second coil. `therefore M_12 =(mu_0N_1N_2a)/l` ....(2) Equations (1) and (2) prove that `M_12=M_21`...(3) From equations (1),(2) and (3) `M_12=M_21=M=(mu_0N_1N_2a)/l` (where `a=pir_1^2`=cross-sectional area of small coil) |
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