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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(1R11R2)
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=B1B2

(considering inward direction to the plane of paper is positive)

Bnet=[μoI4R1μoI4R2]

Bnet=μoI4[1R11R2]

Thus, the direction of the resultant field will be perpendicular and into the plane of page.

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