Q64 / 90 · JEE Main 2024 (08 Apr Shift 2) · Physics

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{a}{2}$ and $2 a$ from axis of the wire is :

  • A
    $1: 4$
  • Correct answer
    B
    $1: 1$
  • C
    $3: 4$
  • D
    $4: 1$

Solution

$\begin{aligned} & \mathrm{B}_1 2 \pi \frac{\mathrm{a}}{2}=\mu_{\mathrm{o}} \frac{\mathrm{I}}{4} \\ & \mathrm{~B}_1=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{a}} \\ & \mathrm{B}_2 2 \pi 2 \mathrm{a}=\mu_{\mathrm{o}} \mathrm{I} \\ & \mathrm{B}_2=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{a}}\end{aligned}$

Hence, the correct answer is option: 'B'.

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