JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
- A \([a / 4,3 a / 2]\)
- B \([a / 4,2 a]\)
- C \([\mathrm{a} / 2,2 \mathrm{a}]\)
- D \([a / 2,3 a]\)
Answer & Solution
Correct Answer
(C) \([\mathrm{a} / 2,2 \mathrm{a}]\)
Step-by-step Solution
Detailed explanation
Maximum possible magnetic field is at the surface \(\begin{aligned} & \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\ & \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}} \end{aligned}\) It can be obtained inside as well as outside the wire…
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