The magnetic field at center O of the arc in figure is
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a
μ0I4π×r[2+π]
b
μ0I2πrπ4+(2-1)
c
μ04π×Ir[2-π]
d
μ04π×Ir2+π4
answer is B.
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Detailed Solution
Magnetic field due to straight wires Bcircular =14μ0I2r=μ0I8r Bstaight =2μ0I4πrcos45°sin90°-sin45° =2μ0I2πr1-12 B=Bcircular +Bstraight B=μ0I2πrπ4+(2-1)