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Q.

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is :

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a

μ0I2πRπ+1

b

μ0I2πRπ+12

c

μ0I2R

d

μ0I2πRπ−12

answer is B.

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Detailed Solution

B→0=B→01+B→02+B→03+B→04​     =μ0i4πRsin900−sin450⊗+μ0i2R⊙+μ0i4πRsin450+sin900⊙=−μ0i4πR1−12+μ0i2R+μ0i4πR12+1⊙​     =μ0i4πR−1+12+2π+12+1⊙=μ0i4πR2+2π⊙​     =μ0i2πR12+π⊙
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