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Question

A V shape bent wire is carrying current $$I$$ and ends of wire extends to infinity. If the magnetic field at $$P$$ can be written as $$K\tan { \left( \cfrac { \alpha }{ 2 } \right)  }$$, then $$K$$ is

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A
μ0I4πd
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B
μ0I2πd
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C
μ0Iπd
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D
2μ0Iπd
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Solution

The correct option is B $$\cfrac { { \mu }_{ 0 }I }{ 2\pi d } $$
 Magnetic field due to any one segment:

$$\quad
B=\cfrac { { \mu  }_{ 0 }i }{ 4\pi (d\sin { \alpha  } ) } \left( \cos { {
0 }^{ o } } +\cos { \left( 180-\alpha  \right)  }  \right) $$

$$=\cfrac
{ { \mu  }_{ 0 }i }{ 4\pi (d\sin { \alpha  } ) } (1-\cos { \alpha  }
)=\cfrac { { \mu  }_{ 0 }I }{ 4\pi d } \tan { \cfrac { \alpha  }{ 2 }  }
$$

Resultant field will be

$${ B }_{ net }=2B=\cfrac { { \mu  }_{
0 }I }{ 2\pi d } \tan { \cfrac { \alpha  }{ 2 }  } \Rightarrow K=\cfrac
{ { \mu  }_{ 0 }I }{ 2\pi d } $$

Physics
NCERT
Standard XII

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