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Two concentric circular coils, one of small radius r, 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 : Let a current `I_2 `flow through the outer circular coil The field at the centre of the coil is `B_2 = mu_0 I_2// 2r_2` . SINCE the other co-axially placed coil has a very small radius, B may be considered constant over its cross-sectional AREA, Hence. `phi_1 = pi r_1^2 B_2 = (mu_0 pi r_1^2)/(2r_2) I_2 = M_12 I_2 ` thus  mutual inductance of solenoid `S_1` with respect to `S_2` `M_12 = (mu_0 pi r_1^2)/(2r_2) ......(i) " but" M_12 = M_21 = (mu_0 pi r_1^2)/(2r_2)` Note that we calculated `M_12`from an approximate value of `phi_1` ,assuming the magnetic field `B_2` to be uniform over the area `pi r_1^2`. However we can accept this value because `r_1 lt lt r_2` ? It would have been difficult to calculate the flux through the BIGGER coil of the non-uniform field due to the current in the SMALLER coil and hence the mutual inductance `M_12` The equality `M_12 = M_21`is HELPFUL. Note also that mutual inductance depends solely on the geometry. |
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