A long wire bent as shown in figure carries current I. If the radius of the semi circular portion is a, the magnetic field at center C is
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a
μ0I4a
b
μ0I4πaπ2+4
c
μ0I4a+μ0I4πa
d
μ0I4πaπ2-4
answer is B.
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Detailed Solution
B1=μ04π×2πIa×12 (due to semicircular part) B2=μ04π×2Ia (due to parallel parts of currents) These two fields are at right angles to each other. Hence, resultant field B=B12+B22=μ0I4πaπ2+4