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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-urially with centres coinciding. Obtain the mutual inductance of the arangement. |
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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_(2)` may be considered CONSTANT over its cross- sectional area, Hence. `Phi_(1)=pi r_(1)^(2)B_(2)=(mu_(0)pir_(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)pir_(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^(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 nonuniform 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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