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 i0, 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 x from the axis when (i) x &lt; a and(ii) a &lt; x &lt; b?

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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 i₀, 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 x from the axis when (i) x < a and(ii) a < x < b?

Option 1 -

$\frac{x^{2}}{a^{2}}$

Option 2 -

$\frac{b^{2} - a^{2}}{x^{2}}$

Option 3 -

$\frac{a^{2}}{x^{2}}$

Option 4 -

$\frac{x^{2}}{b^{2} - a^{2}}$

3 Views | Posted 2 months ago

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

For x < a,

B₁ = $\frac{μ_{0} i_{0} x}{2 π a^{2}}$

For a < x < b,

$B_{2} = \frac{μ_{0} i_{0}}{2 π x}$

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

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