In the given figure, what is the magnetic field induction at point O.
A
μ0l4πr
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B
μ0l4r+μ0l2πr
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C
μol4r+μ0l4πr
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D
μ0l4r−μ0l4πr
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Solution
The correct option is Cμol4r+μ0l4πr Magnetic field due to straight wire above O is zero i.e., B1=0(since, θ=0o) The magnetic field due to semi circular part B2=(μ0nI2r)=μ0(1/2)I2r=μ0I4r The magnetic field due to lower straight portion B3=12μ0I2πr(upward) Net magnetic field B=B1+B2+B3 =0+μ0I4r+μ0I4πr(upwards).