Q.

A wire is bent in the form of a circular arc with a straight portion AB. Magnetic induction at O when current I flowing in the wire, is

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a

μ02r(π-θ+tanθ)

b

μ0I2πr(π+θ-tanθ)

c

μ0I2πr(π-θ+tanθ)

d

μ0I2πr(-tanθ+π-θ)

answer is C.

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Detailed Solution

BAB=μ0I4π(OC)[2sinθ]  But OC=rcosθ or BAB=μ0I2πrtanθ Magnetic field due to circular portion, BAB=μ0I2r2π-2θ2π=μ0I2πr(π-θ) Total magnetic field =μ0I2πrtanθ+μ0I2πr(π-θ)=μ0I2πr[tanθ+π-θ]
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