| Directions: are based on the following paragraph. |
| 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 \[30{}^\text{o}.\] another straight thin wire with steady current\[{{I}_{1}}\]flowing out of the plane of the paper is kept at the origin. |
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| The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is [AIEEE 2009] |
A) zero
B) \[\frac{{{\mu }_{0}}I(b-a)}{24ab}\]
C) \[\frac{{{\mu }_{0}}I}{4\pi }\left[ \frac{b-a}{ab} \right]\]
D) \[\frac{{{\mu }_{0}}I}{4\pi }\left[ 2(b\_a)+\frac{\pi }{3}(a+b) \right]\]
Correct Answer: B
Solution :
[b] \[B=\frac{1}{12}\]of\[\frac{{{\mu }_{0}}I}{2}\left( \frac{1}{a}-\frac{1}{b} \right)\] \[=\frac{{{\mu }_{0}}I(b-a)}{24ab}\].
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