JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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
- A \(\frac{2}{3}\)
- B \(\frac{3}{2}\)
- C \(\frac{1}{2}\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
Let current density be J. \(\therefore\) Applying Ampere's law. \(\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\mu_{0} \mathrm{i} \Rightarrow \mathrm{B}_{\mathrm{A}} 2 \pi \frac{\mathrm{a}}{3}=\mu_{0} \mathrm{J} \pi\left(\frac{\mathrm{a}}{3}\right)^{2}\)…
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