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  • A coaxial cable consists of an inner wire of radius surrounded by an outer shell of inner and outer radii <img alt="' b$ ' and ' $c '" src="https://entrancecorner.codecog

A coaxial cable consists of an inner wire of radius ' a 'surrounded by an outer shell of inner and outer radii ' b$ ' and ' $c 'respectively. The inner wire carries an electric current i_{0}, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance xfrom the axis when (i) x<\mathrm{a}$ and (ii) $\mathrm{a}<x<\mathrm{b} ?
Option: 1 \begin{aligned} & \frac{b^{2}-a^{2}}{x^{2}} \\ \end{aligned}
Option: 2 \frac{x^{2}}{a^{2}} \\
Option: 3 \frac{a^{2}}{x^{2}} \\
Option: 4 \frac{x^{2}}{b^{2}-a^{2}}

Answers (1)

best_answer

 For \: x< a

Only inner wire will contribute to magetic field

By Ampere's law

B_{1} \times 2 \pi x=\mu_{0} i_{enc} =\frac{\mu_{0} i}{\pi a^{2}} \times \pi x^{2}\\

B_{1}= \frac{\mu_{0}ix}{2\pi a^{2}}\\                          .............(1)

For \: a< x< b

Even  here only inner wire will contribute

i_{enc}= i

By Ampere's law

\oint \bar{B}\cdot \bar{dl}= \mu_{0}i_{enc}\\

B_{2}\times2\pi x= \mu_{0}i_{0}\\

B_{2}= \frac{\mu_{0}i}{2\pi x}\\                     ......(2)

\frac{B_{1}}{B_{2}}= \frac{x^{2}}{a^{2}}

Posted by

vishal kumar

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