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  • A long straight wire with a circular cross-section having radius , is carrying a steady current I. The current I is uniformly

A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r$ ( $r<R) from its centre will be :
 

Option: 1

B \propto r^{2}


Option: 2

B \propto r


Option: 3

B \propto \frac{1}{r^{2}}


Option: 4

\mathrm{B} \propto \frac{1}{\mathrm{r}}


Answers (1)

best_answer


By Ampere's law,
\mathrm{\oint \bar{B}. d\bar{l}= \mu _{0}\,I_{enc}}
                \mathrm{= \mu _{0}\,J\,A_{enc}}

\mathrm{B\int dl\cos 0^{\circ}= \frac{\mu _{0}I}{\pi R^{2}}\times \pi r^{2}}
\mathrm{B\left ( 2\pi r \right )= \frac{\mu _{0}\, \, I\, r^{2}}{R^{2}}}
\mathrm{B= \frac{\mu _{0}\, \, I\, r}{2\pi R^{2}}}

Here pt .p lies at \mathrm{r< R}

\mathrm{\therefore \: B\, \alpha \, r\quad\left ( for\; r< R \right )}

The correct option is (2)

 

Posted by

rishi.raj

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