Two wires PQ and QR, carry equal currents i as shown in figure. One end of both the wires extends to infinity $∠ PQR = θ$ . The magnitude of the magnetic field at O on the bisector angle of these two wires at a distance r from point Q, is

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$\frac{μ_{0}}{4 π} \frac{i}{r} \sin \frac{θ}{2}$

$\frac{μ_{0}}{4 π} \frac{i}{r} \tan \frac{θ}{2}$

$\frac{μ_{0}}{4 π} \frac{i}{r} \cot θ$

$\frac{μ_{0}}{2 π} \frac{i}{r} \frac{(1 + \cos θ / 2)}{(\sin θ / 2)}$

answer is D.

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Detailed Solution

Length of perpendicular to O from PQ, QR will be $a = r \sin \frac{θ}{2}$
Magnetic field induction at O due to current through PQ and QR is

$B = \frac{μ_{0}}{4 π} \frac{i}{a} [\sin (90^{0} - \frac{θ}{2}) + \sin 90^{0}] \times 2$

( field due to a straight wire)

$= \frac{μ_{0}}{2 π} \frac{i}{r \sin \frac{θ}{2}} (\cos \frac{θ}{2} + 1) = \frac{μ_{0}}{2 π} \frac{i}{r} \frac{(1 + \cos \frac{θ}{2})}{\sin \frac{θ}{2}}$

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