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Question:

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30o. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin. The magnitude of the magnetic field due to the loop ABCD at the origin (O) is:?

zero

μ0 I(b-a)/24ab

μ0 I/4π [2(b-a)+π/3(a+b)].

μ0 I/4π [(b-a)/ab]

Solution:

The net magnetic at the origin O can be given as the sum of the field due to each arm of the loop.
So, the field due to all loop is:
B = BAB + BBC + BCD + BDA
B = 0 + μ0 I/4πa × π/6 + 0 - μ0 I/4πb × π/6
B = μ0 I/24ab(b-a)