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As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If $$OP = OQ = 4$$ cm, and the magnitude of the magnetic field at O is $$10^{-4}$$ T, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be $$\left(\mu_0 = 4\pi \times 10^{-7} NA^{-2}\right)$$:
Axial segments: $$B_{LP} = B_{MQ} = 0$$
Perpendicular segments at origin: $$\vec{B}_{PS} = \frac{\mu_0 I}{4\pi d} (-\hat{k})$$
$$\vec{B}_{QN} = \frac{\mu_0 I}{4\pi d} (-\hat{k})$$
Total magnetic field:
$$\vec{B}_{\text{net}} = \vec{B}_{PS} + \vec{B}_{QN} = \frac{\mu_0 I}{2\pi d} (-\hat{k})$$
$$10^{-4} = \frac{(4\pi \times 10^{-7}) \cdot I}{2\pi \times 0.04} = 5 \times 10^{-6} I \implies I = 20\text{ A}$$
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