A long straight wire of radius a… - NEET UG 2016 Physics PYQ

NEET UG 2016 — Physics Electromagnetism

2016mcqmedium

A long straight wire of radius $\text{a}$ carries a steady current $\text{I}$ . the current is uniformly distributed over its cross-section. The ratio of the magnetic fields $\text{B}$ and $\text{B'}$ , at radial distances $\frac{a}{2}$ and $2a$ respectively, from the axis of the wire is:

Official previous-year question

Held on 30 Apr 2016 · Verified 9 Jul 2026.

Options

A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
1
D
$4$

Did you get this right?

Solution

$\frac{Inside the wire}{By amper{e}^{'}s law}$

$\int \vec{B}.\vec{dl}={\mu }_{0}({i}_{enclosed})$

$\int B.dl\mathrm{cos}0={\mu }_{0}(\frac{I}{\pi {a}^{2}}.\pi {(\frac{a}{2})}^{2})$

$B\int dl={\mu }_{0}\frac{I}{4}$

$B(2\pi (\frac{a}{2}))=\frac{{\mu }_{0}I}{4}$

$B=\frac{{\mu }_{0}I}{4\pi a}$

Outside the wire,

${B}^{'}=\frac{{\mu }_{0}I}{2\pi r}=\frac{{\mu }_{0}I}{2\pi (2a)}=\frac{{\mu }_{0}I}{4\pi a}$

So, $\frac{B}{{B}^{'}}=1$ .

Your note

NEET UG Electromagnetism in other years

Electromagnetism 2026 Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2023 Electromagnetism 2022 Electromagnetism 2021 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2015 Electromagnetism 2014

More NEET UG Electromagnetism questions

Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:

2026medium

A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is :

2025medium

Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :

2025medium

The electric potential at distance r from a point charge q is V = q/(4πε₀r). The electric field E is:

2025easy

The current passing through the battery in the given circuit, is : ![NEET 2025 Physics, Current Electricity — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Current_Electricity/682c9a94c2281abe1696a0a6/question_1__q_682c9a94c2281abe1696a0a6__cdn-question-pool.getmarks.app__yKwOmrdZMSd9Ml6L9jUq5lCIlM2f2z32bgW-sqY2ViU.original.fullsiz__b2faee45bc_final_ppt_sync.png)

2025hard

View all Electromagnetism questions

Keep practicing Electromagnetism

Work through every NEET UG Electromagnetism PYQ, year by year.

Browse all Physics topics

Solution

$\frac{Inside the wire}{By amper{e}^{'}s law}$

$\int \vec{B}.\vec{dl}={\mu }_{0}({i}_{enclosed})$

$\int B.dl\mathrm{cos}0={\mu }_{0}(\frac{I}{\pi {a}^{2}}.\pi {(\frac{a}{2})}^{2})$

$B\int dl={\mu }_{0}\frac{I}{4}$

$B(2\pi (\frac{a}{2}))=\frac{{\mu }_{0}I}{4}$

$B=\frac{{\mu }_{0}I}{4\pi a}$

Outside the wire,

${B}^{'}=\frac{{\mu }_{0}I}{2\pi r}=\frac{{\mu }_{0}I}{2\pi (2a)}=\frac{{\mu }_{0}I}{4\pi a}$

So, $\frac{B}{{B}^{'}}=1$ .

Options

Did you get this right?

Solution

Your note

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

Options

Did you get this right?

Solution

Your note

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