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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2020
  • Hard
  • MCQ

A long, straight wire of radius \(a\) carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance \(\frac{a}{3}\) and \(2 a,\) respectively from the axis of the wire is

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Answer & Solution

Correct Answer

(A) \(\frac{2}{3}\)

Step-by-step Solution

Detailed explanation

Let current density be J.

\(\therefore\) Applying Ampere's law.

\(\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\mu_{0} \mathrm{i} \Rightarrow \mathrm{B}_{\mathrm{A}} 2 \pi \frac{\mathrm{a}}{3}=\mu_{0} \mathrm{J} \pi\left(\frac{\mathrm{a}}{3}\right)^{2}\)

\(\therefore B_{A}=\frac{\mu_{0} J a}{6}\)

Similarly, \(B_{B}=\frac{\mu_{0} J a}{4}\)

\(\frac{B_{A}}{B_{B}}=\frac{\mu_{0} J a \times 4}{\mu_{0} J 6 a}=\frac{2}{3}\)

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