JEE Main · 2007 · 29 Apr 2007 · Offline · Magnetic Effects of Current

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Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current

I_{1}

and COD carries a current

I_{2}

. The magnetic field on a point lying at a distance '

d

' from

O

, in a direction perpendicular to the plane of the wires

A O B

and

C O D

, will be given by

(A) $\frac{μ _{0}}{2 π} (\frac{I _{1} + I _{2}}{d})^{1/2}$
(B) $\frac{μ _{0}}{2 π d} (I_{1}^{2} + I_{2}^{2})^{1/2}$ ✓ Correct
(C) $\frac{μ _{0}}{2 π d} (I_{1} + I_{2})$
(D) $\frac{μ _{0}}{2 π d} (I_{1}^{2} + I_{2}^{2})$

Answer: (B)

Solution

Each straight wire sets up a field at the point a distance

d

away of magnitude

\frac{μ _{0} I}{2 π d}

. Because the two wires cross at right angles, the point lying off their plane receives two contributions that are themselves perpendicular, one from each current. Adding them as perpendicular vectors gives a resultant of

\frac{μ _{0}}{2 π d} I_{1}^{2} + I_{2}^{2}

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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JEE Main · 2007 · 29 Apr 2007 · Offline · Magnetic Effects of Current

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