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Q.

The magnetic field at center ‘O’ of the arc in figure is

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a

μoI4π×r2+π

b

μoI2πrπ4+2−1

c

μ04π×Ir2−π

d

μ04π×Ir2+π4

answer is B.

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Detailed Solution

Magnetic field due to straight wires and quarter circle is given byBcircular=14μ0I2r=μ0I8r Bstraight=2μ0I4π(rcos450)sin900−sin450 =2μ0I2πr1−12  Hence, Bnet = μ0I8r+2μ0I2πr1−12 = μoI2πrπ4+2−1
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