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Question

The magnetic induction at the center O in the figure is,

A
μ0I4(1R11R2)
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B
μ0I4(1R1+1R2)
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C
μ0I4(R1R2)
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D
μ0I4(R1+R2)
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Solution

The correct option is A μ0I4(1R1−1R2)Magnetic field due to the two straight wire segments at O is zero because point O is lying on the axis of wire. So magnetic field at point O will be produced due to semicircular segments only. Thus, the magnetic field due to the inner semicircle of radius R1 is given by, B1=μ0I4R1 Now using right- hand thumb rule for the direction of magnetic field ∴B1=μ0I4R1 ⨂ ⨂ indicates the direction of magnetic field perpendicularly inwards to the plane of paper. Similarly, the magnetic field due to the outer semicircle of radius R2 is given by, B2=μ0I4R2 ∴B2=μ0I4R2 ⨀ So, the net magnetic field at point O will be ∣∣−−→Bnet∣∣=∣∣−→B1∣∣−∣∣−→B2∣∣ (considering inward direction to the plane of paper is positive) Bnet=[μoI4R1−μoI4R2] ∴Bnet=μoI4[1R1−1R2] Thus, the direction of the resultant field will be perpendicular and into the plane of page.

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