A long wire carrying current $i$ is bent to form a plane angle $θ$ . Find the magnetic field $B$ at a point on the bisector of this angle situated at a distance $x$ from the vertex is written in the form of $k cot \frac{θ}{4}$ Tesla. Then find the value of $k$ .

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Solution

$θ_{1}, θ_{2}$ with the end of wire is written as

$B = \frac{μ_{0} I}{4 π a} [sin θ_{1} + sin θ_{2}]$

In this case :
$\begin{matrix} a = sin (\frac{θ}{2}) θ_{1} = \frac{θ}{2} a n d θ_{2} = 90 \end{matrix}$

therefore we get magnetic field due to one wire
$B_{l o w e r} = \frac{μ_{0} I}{4 π a} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1 + sin (\frac{θ}{2})}{x sin (\frac{θ}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$B_{u p p e r}$ also same

therefore Net field

$B_{n e t} = B_{l o w e r} + B_{u p p e r}$
implies that $2 \times \frac{μ_{0} I}{4 π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1 + sin (\frac{θ}{2})}{x sin (\frac{θ}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$\begin{matrix} \frac{μ_{0} I}{2 π x} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1 + sin (\frac{θ}{2})}{sin (\frac{θ}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ \frac{μ_{0} I}{2 π x} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{{sin (\frac{θ}{4}) + cos (\frac{θ}{4})}^{2}}{2 sin (\frac{θ}{4}) cos (\frac{θ}{4})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \end{matrix}$

Now dividing numerator and dinominator by ${cos}^{2} (\frac{θ}{4})$
$\begin{matrix} \frac{μ_{0} I}{4 π x} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{{1 + tan (\frac{θ}{4})}^{2}}{tan (\frac{θ}{4})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \frac{μ_{0} I}{4 π x} [{1 + tan (\frac{θ}{4})}^{2}] cot (\frac{θ}{4}) \end{matrix}$
Given
$B = K cot (\frac{θ}{4})$

And Putting the value from above expression
$K = \frac{μ_{0} I}{4 π x} [{1 + tan (\frac{θ}{4})}^{2}]$

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