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
Pick an option to check your answer.
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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