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A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin . The magnitude of the magnetic field (B) due to the loop `ABCD` at the origin (o) is :A. `(mu_(0)l(b-a))/(24ab)`B. `(mu_(0)iI)/(4pi)[(b-a)/(ab)]`C. `(mu_(0)I)/(4pi)[2(b-a)+pi/3(a+b)]`D. zero |
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Answer» Correct Answer - 1 Magnetic field due to loop `ABCD` `(mu_(0)I)/(4pi)(pi/6)xx[1/a-1/b] =(u_(0)I)/24[(b-a)/(ab)]` |
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