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Q.

ABCD is a square of side 4R and EFGH is a square loop of side 2R. The loop carries a current I. The magnetic field at the centre O of the loop

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a

μ0l2πR⊙

b

μoI2πR⊗

c

μ0l2πR⊗

d

μ0I2πR⊙

answer is D.

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Detailed Solution

4×μ0I4π⋅sin⁡450+sin⁡4501R−12R =μ0Iπ[2]12R=μ0I2πR⊙
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