First slide

Magnetic field due to a moving charge

Question

Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that 0 is their common point for the two. The wires carry $I_{1} a n d I_{2}$ currents, respectively. Point P is lying at distance d from 0 along a direction perpendicular to the plane containing the wires. The magnetic field at the point P will be

Easy

A

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

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

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

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

Solution

The magnetic field at the point P, at a perpendicular distance d from 0 in a direction perpendicular to the plane ABCD due to currents through AOB and COD 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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