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 AOB and COD, will be given by

  • Option 1)

    \frac{\mu _{0}}{2\pi d}(I_{1}\, ^{2}+I_{2}\, ^{2})

  • Option 2)

    \frac{\mu _{0}}{2\pi }(\frac{I_{1}+I_{2}}{d})^{\frac{1}{2}}

  • Option 3)

    \frac{\mu _{0}}{2\pi d}(I_{1}^{2}+I_{2}^{2})^{\frac{1}{2}}

  • Option 4)

    \frac{\mu _{0}}{2\pi d}(I_{1}+I_{2})

 

Answers (1)

As we learnt in

For Infinite Length -

\phi_{1}=\phi_{2}=90^{\circ}

B=\frac{\mu_{o}}{4\pi}\:\frac{2i}{r}\:

- wherein

The field at the same point at the same distance from the mutually perpendicular wires carrying current will be having the same magnitude but in perpendicular directions.

 B=\sqrt{B^2_1 + B^2_2}

B_1=\frac{\mu_0I_1}{2\pi d}

B_2=\frac{\mu_0 I_2}{2\pi d}

B_1=\frac{\mu_0}{2\pi d}(I^2_1+I^2_1)^{\frac{1}{2}}

 

 


Option 1)

\frac{\mu _{0}}{2\pi d}(I_{1}\, ^{2}+I_{2}\, ^{2})

Incorrect

Option 2)

\frac{\mu _{0}}{2\pi }(\frac{I_{1}+I_{2}}{d})^{\frac{1}{2}}

Incorrect

Option 3)

\frac{\mu _{0}}{2\pi d}(I_{1}^{2}+I_{2}^{2})^{\frac{1}{2}}

Correct

Option 4)

\frac{\mu _{0}}{2\pi d}(I_{1}+I_{2})

Incorrect

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