Two straight long conductors AOB and COD are | vedclass

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(C) The magnetic field due to a long straight wire at a distance $a$ is given by $B = \frac{\mu_0 i}{2\pi a}$.For the conductor $AOB$ carrying current

Vedclass · Accepted Answer

$\frac{\mu_0}{2\pi a}(i_1^2 + i_2^2)^{1/2}$

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(C) The magnetic field due to a long straight wire at a distance $a$ is given by $B = \frac{\mu_0 i}{2\pi a}$.For the conductor $AOB$ carrying current $i_1$,the magnetic field at point $P$ is $B_1 = \frac{\mu_0 i_1}{2\pi a}$.For the conductor $COD$ carrying current $i_2$,the magnetic field at point $P$ is $B_2 = \frac{\mu_0 i_2}{2\pi a}$.Since the conductors are perpendicular,the magnetic field vectors $B_1$ and $B_2$ are also perpendicular to each other.The net magnetic field at point $P$ is $B_{net} = \sqrt{B_1^2 + B_2^2}$.Substituting the values,$B_{net} = \sqrt{\left(\frac{\mu_0 i_1}{2\pi a}\right)^2 + \left(\frac{\mu_0 i_2}{2\pi a}\right)^2}$.$B_{net} = \frac{\mu_0}{2\pi a} \sqrt{i_1^2 + i_2^2} = \frac{\mu_0}{2\pi a}(i_1^2 + i_2^2)^{1/2}$.

List-$I$	List-$II$
$(A)$ Fleming's left hand rule	$(i)$ Direction of induced current
$(B)$ Right hand thumb rule	(ii) Magnitude and direction of magnetic induction
$(C)$ Biot-Savart law	(iii) Direction of force due to magnetic induction
$(D)$ Fleming's right hand rule	(iv) Direction of magnetic lines due to current

Two straight long conductors $AOB$ and $COD$ are perpendicular to each other and carry currents $i_1$ and $i_2$. The magnitude of the magnetic induction at a point $P$ at a distance $a$ from the point $O$ in a direction perpendicular to the plane $ACBD$ is

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