Two identical long conducting wires AOB and C | vedclass

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(C) The magnetic field due to a long straight wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{2\pi d}$.For wire $AOB$ carrying cu

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The magnetic field due to a long straight wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{2\pi d}$.For wire $AOB$ carrying current $I_1$,the magnetic field at point $P$ (at distance $d$ along the perpendicular axis) is $B_1 = \frac{\mu_0 I_1}{2\pi d}$.For wire $COD$ carrying current $I_2$,the magnetic field at point $P$ is $B_2 = \frac{\mu_0 I_2}{2\pi d}$.Since the wires are perpendicular to each other,the magnetic field vectors $\vec{B}_1$ and $\vec{B}_2$ at point $P$ are also perpendicular to each other.The resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.Substituting the values: $B = \sqrt{\left( \frac{\mu_0 I_1}{2\pi d} \right)^2 + \left( \frac{\mu_0 I_2}{2\pi d} \right)^2}$.$B = \frac{\mu_0}{2\pi d} \sqrt{I_1^2 + I_2^2} = \frac{\mu_0}{2\pi d} (I_1^2 + I_2^2)^{1/2}$.

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