JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2021
- Hard
- MCQ
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 is, 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 \((ii)\) \({a}<{x}<{b}\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{{x}^{2}}{{a}^{2}}\)
Step-by-step Solution
Detailed explanation
when \({x}<{a}\)
\({B}_{1}(2 \pi {x})=\mu_{\circ}\left(\frac{{i}_{\circ}}{\pi {a}^{2}}\right) \pi {x}^{2}\)
\({B}(2 \pi {x})=\frac{\mu_{\circ} {i}_{\circ} {x}^{2}}{{a}^{2}}\)
\({B}_{1}=\frac{\mu_{\circ} {i}_{\circ} {x}}{2 \pi {a}^{2}}...(1)\)
when \({a}<{x}<{b}\)
\({B}_{2}(2 \pi {x})=\mu_{0} {i}_{0}\)
\({B}_{2}=\frac{\mu_{{o}} {i}_{\circ}}{2 \pi {x}}...(2)\)
\(\frac{{B}_{1}}{{B}_{2}}=\frac{\mu_{\circ} {i}_{\circ} \frac{{x}}{2 \pi {a}^{2}}}{\frac{\mu_{\circ} {i}_{\circ}}{2 \pi {x}}}=\frac{{x}^{2}}{{a}^{2}}\)
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