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Two identical long conducting wires AOB and COD 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 I1 and I2 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 :

  • Option 1)

    \frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}/\text{I}_{2} \right )

  • Option 2)

    \frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}+\text{I}_{2} \right )

  • Option 3)

    \frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}^{2}-\text{I}_{2} ^{2}\right )

  • Option 4)

    \frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}^{2}+\text{I}_{2}^{2} \right )^{1/2}

 

Answers (1)

 

Resultant field B=

\sqrt{B_{1}^{2}+B_{2}^{2}}\\ B_{1}=\frac{\mu _{oI_{1}}}{2\pi d} and B_{2}=\frac{\mu _{oI_{2}}}{2\pi d}\\ B=\frac{\mu _{o}}{2\pi d}(I_{1}^{2}+I_{2}^{2})^{1/2}


Option 1)

\frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}/\text{I}_{2} \right )

This solution is incorrect 

Option 2)

\frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}+\text{I}_{2} \right )

This solution is incorrect 

Option 3)

\frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}^{2}-\text{I}_{2} ^{2}\right )

This solution is incorrect 

Option 4)

\frac{\mu_{\text{o}}}{2\pi\text{d}}\left ( \text{I}_{1}^{2}+\text{I}_{2}^{2} \right )^{1/2}

This solution is correct 

Posted by

Sabhrant Ambastha

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