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A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is,

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μ0i2R, inward

Zero

3μ0i/32R, outward

3μ0i/32R, inward

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detailed solution

Correct option is B

Net magnetic field at point 'P'Bnet=B1→+B2→Here B1→ and B2→ are equal in magnitude and opposite in direction.Hence,Bnet=B1-B2i1=iθ2π⇒B1=μ0i12R2π-θ2πi2=i2π-θ2π⇒B2=μ0i22Rθ2πBnet=B1-B2=0

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