A thick current carrying cable… - NEET UG 2021 Physics PYQ

NEET UG 2021 — Physics Electromagnetism

2021mcqeasy

A thick current carrying cable of radius $R$ carries current $I$ uniformly distributed across its cross-section. The variation of magnetic field $B(r)$ due to the cable with the distance $r$ from the axis of the cable is represented by:

Official previous-year question

Held on 30 Apr 2021 · Verified 9 Jul 2026.

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Solution

NEET 2021 Physics, Magnetic Effects of Current — solution figure 1

For inside part

$J=\frac{I}{\pi {R}^{2}}$

$\int \vec{B}\cdot \vec{dl}={\mu }_{0}{I}_{inclosed}\ldots (i)$

$B2\pi r={\mu }_{0}(\frac{I}{\pi {R}^{2}})\times \pi {r}^{2}$

$B2\pi r={\mu }_{0}I\frac{{r}^{2}}{{R}^{2}}$

$B\propto r$

For outside part

$\int \vec{B}\cdot \vec{dl}={\mu }_{0}{I}_{inclosed}$

$B(2\pi r)={\mu }_{0}I$

$\therefore B\propto \frac{1}{r}$

NEET 2021 Physics, Magnetic Effects of Current — solution figure 2

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Solution

NEET 2021 Physics, Magnetic Effects of Current — solution figure 1

For inside part

$J=\frac{I}{\pi {R}^{2}}$

$\int \vec{B}\cdot \vec{dl}={\mu }_{0}{I}_{inclosed}\ldots (i)$

$B2\pi r={\mu }_{0}(\frac{I}{\pi {R}^{2}})\times \pi {r}^{2}$

$B2\pi r={\mu }_{0}I\frac{{r}^{2}}{{R}^{2}}$

$B\propto r$

For outside part

$\int \vec{B}\cdot \vec{dl}={\mu }_{0}{I}_{inclosed}$

$B(2\pi r)={\mu }_{0}I$

$\therefore B\propto \frac{1}{r}$

NEET 2021 Physics, Magnetic Effects of Current — solution figure 2

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Options

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Solution

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NEET UG Electromagnetism in other years

More NEET UG Electromagnetism questions

Other Physics topics for NEET UG

Mechanics

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Electromagnetism