Verified Solution • 2025–26 Session moderate

QJEEM25Ph453a: JEE Main 2025 Physics - Class 12 JEEM25Ph453a Solution: A long straight wire of a circular cross-section with radius '…

A long straight wire of a circular cross-section with radius ' a ' carries a steady current I . The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance rr from the centre of the wire is given by

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/8b09a32b4e8aa0a29bb341b53b8384f77172f23397fd4764b52a20c38c01ac19.webp
  • B https://ik.imagekit.io/novatra/photos/62ca4bbd562ce91c06e6a5290d561f8bbb5f1c55b969c7edf3ae12cdc699fa1b.webp
  • C https://ik.imagekit.io/novatra/photos/c0f69c13848ba343b49b3fcdfc5c66e07e9217bee832fdfec45356d9330e1004.webp
  • D https://ik.imagekit.io/novatra/photos/2fb86c814e83ada4ee0c774e99104d87caf1242aae59ea8480f2721d304a30d8.webp

Step-by-Step Verified Solution

JEE Main 2025 (Online) 24th January Evening Shift Physics - Magnetic Effect of Current Question 3 English Explanation

amp;Bin =μ0( J‾×r‾)2amp; Bin ∝ramp; Bout =μ0I2πramp; Bnet ∝1r\begin{aligned}& \mathrm{B}_{\text {in }}=\frac{\mu_0(\overline{\mathrm{~J}} \times \overline{\mathrm{r}})}{2} \\& \mathrm{~B}_{\text {in }} \propto \mathrm{r} \\& \mathrm{~B}_{\text {out }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\& \mathrm{~B}_{\text {net }} \propto \frac{1}{\mathrm{r}}\end{aligned}

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