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Two circular coils, one of smaller radius r_(1) and the other of very large radius r_(2) are placed co-axially with centres coinciding. Obtain the mutal inductance of the arrangement. |
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Answer» Solution :Suppose a current `I_(2)` flows through the outer circular coil. The field at the centre of the coil is `B_(2)= (mu_(0)I_(2))/(2r_(2))`  The second co-axially placed coil has very small radius. So `B_(2)` may be considered constant over its cross-sectional area Now, `phi_(1)= pi r_(1)^(2) B_(2)= pi r_(1)^(2) ((mu_(0)I_(2))/(2r_(2)))` or `phi_(1)= (mu_(0)pi r_(1)^(2))/(2r_(2)) I_(2)` Comparing with `PHI= M_(12)I_(2)`, we get `M_(12)= (mu_(0)pi r_(1)^(2))/(2r_(2))` Also, `M_(21)= M_(12)= (mu_(0)pi r_(1)^(2))/(2r_(2))` It WOULD havebeen 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 helpul. Note also that mutual inductance depends solely on the geometry. |
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