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Two concentric circular coil,s 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. |
Answer» Let a current `I_(2)` flow through the outer circuler 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_(2)=pir_1^(2)B_(2)` `=(mu_(0)pir_(1)^(2))/(2r_(2))I_(2)` `=M_(12)I_(2)` Thus, `M_(12)==(mu_(0)pir_(1)^(2))/(2r_(2))` From equation `M_(12)=M_21=(mu_(0)pir_(1)^(2))/(2r_(2))` Note that we calculate `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)`. |
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