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Magnetics question

2009 · Shift 0 · Q58
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Magnetics question

2009 · Shift 0 · Q58

JEE MainPhysicsMagneticsMCQ+4 / −1
A current loop ABCDABCDABCD is held fixed on the plane of the paper as shown in the figure. The arcs BCBCBC(radius =b= b=b) and DADADA(radius =a=a=a) of the loop are joined by two straight wires ABABAB and CDCDCD. A steady current III is flowing in the loop. Angle made by ABABAB and CDCDCD at the origin OOO is 30∘.{30^ \circ }.30∘. Another straight thin wire steady current I1{I_1}I1​ flowing out of the plane of the paper is kept at the origin. AIEEE 2009 Physics - Magnetic Effect of Current Question 195 English The magnitude of the magnetic field (B)(B)(B) due to the loop ABCDABCDABCD at the origin (O)(O)(O) is :
  1. A
    μ0I(b−a)24ab{{{\mu _0}I\left( {b - a} \right)} \over {24ab}}24abμ0​I(b−a)​
  2. B
    μ0I4π[b−aab]{{{\mu _0}I} \over {4\pi }}\left[ {{{b - a} \over {ab}}} \right]4πμ0​I​[abb−a​]
  3. C
    μ0I4π[2(b−a)+π3(a+b)]{{{\mu _0}I} \over {4\pi }}\left[ {2\left( {b - a} \right) + \frac{\pi }{3}\left( {a + b} \right)} \right]4πμ0​I​[2(b−a)+3π​(a+b)]
  4. D
    zero
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