If a wire carrying current "I" is bent to two arms making right angle between them, then the magnetic field intensity B at the distance "a" from both arms is
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Given that,
Current =i
Angle =α
Magnetic field =B
Now, magnetic field at a point a from current carrying conductor making angle θ
1
and θ
2
with the ends of the wire is
B=
4πa
μ
0
I
(sinθ
1
+sinθ
2
)
Now, a=xsin
2
θ
θ
1
=
2
θ
θ
2
=90
0
Now, the magnetic field
B
lower
=
4π
μ
0
I
⎣
⎢
⎢
⎢
⎡
xsin(
2
θ
)
1+sin(
2
θ
)
⎦
⎥
⎥
⎥
⎤
And B
upper
is also same
No, if the current was flowing anticlockwise both these field would point upward outside the page
Now, net magnetic field
B=B
lower
+B
upper
B=2×
4π
μ
0
I
⎣
⎢
⎢
⎢
⎡
xsin(
2
θ
)
1+sin(
2
θ
)
⎦
⎥
⎥
⎥
⎤
B=
2πx
μ
0
I
⎣
⎢
⎢
⎢
⎡
sin
2
θ
1+sin(
2
θ
)
⎦
⎥
⎥
⎥
⎤
B=
2πx
μ
0
I
⎣
⎢
⎢
⎢
⎢
⎡
2sin(
4
θ
)cos(
4
θ
)
{sin(
4
θ
)+cos(
4
θ
)}
2
⎦
⎥
⎥
⎥
⎥
⎤
Now, dividing numerator and denominator by cos
2
4
θ
B=
4πx
μ
0
I
⎣
⎢
⎢
⎢
⎢
⎡
tan
4
θ
(1+tan(
4
θ
))
2
⎦
⎥
⎥
⎥
⎥
⎤
B=
4πx
μ
0
I
[(1+tan(
4
θ
))
2
]cot(
4
θ
)
Now, the magnetic field is
B=Kcot(
4
θ
)
Where, K=
4πx
μ
0
I
[{1+tan(
4
θ
)}
2
]
Hence, the magnetic field is B=Kcot(
4
θ
)
Explanation:
hope it helps you.........