
B0=(B0)1+(B0)2+(B0)3+(B0)4
4πRμ0i[sin90∘−sin45∘]⨂+2Rμ0i⨀+ 4πRμ0i(sin45∘+sin90∘)⨀
−4πRμ0i[1−21]+2Rμ0i+4πRμ0i[21+1]⨀
=4πRμ0i[−1+21+2π+21+1]⨀=4πRμ0i[2+2π]⨀=2πRμ0i[21+π]⨀
JEE Main 2020 — Physics Electromagnetism
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 filed at the center of circle is:

Held on 8 Jan 2020 · Verified 6 Jul 2026.
2πRμ0I(π+21)
2πRμ0I(π−21)
2πRμ0I(π+1)
2Rμ0I
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