An infinitely long wire bent into shape as shown in figure carries a steady current I. If the curved wire is along the curve $y^{2} = 4 a x$ , then magnetic field at point $P (a, 0)$ , would be $\frac{μ_{0} l}{4 a} (\frac{π + x}{2 π})$

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Detailed Solution

Apply Biot Savart’s law.

$B = B_{1} + B_{2}$

$B_{1}$ (Due to semi-infinite straight wire) $= \frac{μ_{0} l}{4 π a}$

$B_{2}$ (Due to parabolic portion) $= \frac{μ_{0} l}{8 a}$

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

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