Two identical conducting wires AOB and COD ar | vedclass

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(C) The magnetic field produced by a long straight current-carrying wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.Since the wires $

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$\frac{\mu_0}{2\pi d} (I_1^2 + I_2^2)^{\frac{1}{2}}$

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(C) The magnetic field produced by a long straight current-carrying wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.Since the wires $AOB$ and $COD$ are placed at right angles to each other,the magnetic fields $B_1$ and $B_2$ produced by them at a point $P$ (at distance $d$ from $O$ along the normal to the plane) will also be perpendicular to each other.Thus,the magnitude of the resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.Substituting the values,$B_1 = \frac{\mu_0 I_1}{2 \pi d}$ and $B_2 = \frac{\mu_0 I_2}{2 \pi d}$.Therefore,$B = \sqrt{\left( \frac{\mu_0 I_1}{2 \pi d} \right)^2 + \left( \frac{\mu_0 I_2}{2 \pi d} \right)^2}$.$B = \frac{\mu_0}{2 \pi d} \sqrt{I_1^2 + I_2^2} = \frac{\mu_0}{2 \pi d} (I_1^2 + I_2^2)^{1/2}$.

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