An infinitely long conductor PQR is bent to form a right angle, as shown. A current I flows through PQR. The magnetic field due to this current at point M is B₁. Now, another infinitely long straight conductor QS is connected to Q so that the current is $\frac{I}{2}$ in QR as well as in QS, and the current in PQ remains unchanged. The magnetic field at M is now B₂. The ratio B₁/B₂ is given by

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answer is C.

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Detailed Solution

The magnetic field at M due to Q R is zero.

Magnetic field at M due to PQ
$B_{1} = \frac{μ_{0} i}{4 π d}$ ;

The magnetic field at M due to PQ and QS
$B_{2} = \frac{μ_{0} i}{4 πd} + \frac{μ_{0} \frac{i}{2}}{4 πd}, B_{2} = \frac{3}{2} \frac{μ_{0} i}{4 π d'} H e n c e, \frac{B_{1}}{B_{2}} = \frac{2}{3}$

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