Two identical long conducting wires and are placed at right angle to each other, with one above other such that is their common point for the two. The wires carry and currents, respectively. Point is lying at distance from along a direction perpen

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Solution

Q.
Two identical long conducting wires $A O B$ and $C O D$ are placed at right angle to each other, with one above other such that $O$ is their common point for the two. The wires carry $I_{1}$ and $I_{2}$ currents, respectively. Point $P$ is lying at distance $d$ from $O$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point $P$ will be [2014]

1 $\frac{μ_{0}}{2 π d} (\frac{I_{1}}{I_{2}})$

2 $\frac{μ_{0}}{2 π d} (I_{1} + I_{2})$

3 $\frac{μ_{0}}{2 π d} (I_{1}^{2} - I_{2}^{2})$

4 $\frac{μ_{0}}{2 π d} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

Ans.
(4)

The magnetic field at the point $P$ , at a perpendicular distance $d$ from $O$ in a direction perpendicular to the plane $A B C D$ due to currents through $A O B$ and $C O D$ are perpendicular to each other. Hence,

$B = {(B_{1}^{2} + B_{2}^{2})}^{1 / 2}$

$= [{(\frac{μ_{0}}{4 π} \frac{2 I_{1}}{d})}^{2} + {(\frac{μ_{0}}{4 π} \frac{2 I_{2}}{d})}^{2}]^{1 / 2}$

$= \frac{μ_{0}}{2 π d} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

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