ROSHAN MUJEEB
Last Activity: 4 Years ago
θ1,θ2with the end of wire is written as
B=4πaμ0I[sinθ1+sinθ2]
In this case :
a=sin(2θ)θ1=2θandθ2=90
therefore we get magnetic field due to one wire
Blower=4πaμ0I⎣⎢⎢⎢⎡xsin(2θ)1+sin(2θ)⎦⎥⎥⎥⎤
Bupperalso same
therefore Net field
Bnet=Blower+Bupper
implies that2×4πμ0I⎣⎢⎢⎢⎡xsin(2θ)1+sin(2θ)⎦⎥⎥⎥⎤
2πxμ0I⎣⎢⎢⎢⎡sin(2θ)1+sin(2θ)⎦⎥⎥⎥⎤2πxμ0I⎣⎢⎢⎢⎢⎡2sin(4θ)cos(4θ){sin(4θ)+cos(4θ)}2⎦⎥⎥⎥⎥⎤
Now dividing numerator and dinominator bycos2(4θ)
4πxμ0I⎣⎢⎢⎢⎢⎡tan(4θ){1+tan(4θ)}2⎦⎥⎥⎥⎥⎤4πxμ0I[{1+tan(4θ)}2]cot(4θ)
Given
B=Kcot(4θ)
And Putting the value from above expression
K=4πxμ0I[{1+tan(4θ)}2]