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Magnetics question

2022 · 28 Jun · Shift 1 · Q52

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2022 · 28 Jun · Shift 1 · Q52

Magnetics question

2022 · 28 Jun · Shift 1 · Q52

JEE MainPhysicsMagneticsMCQ+4 / −1

An infinitely long hollow conducting cylinder with radius R carries a uniform current along its surface. Choose the correct representation of magnetic field (B) as a function of radial distance (r) from the axis of cylinder.

A
B
C
D

View written solutionFree

Correct answer: D

Understand the current distribution

An infinitely long hollow conducting cylinder of radius $R$ carries current uniformly along its surface and parallel to its axis.

So the current is confined only at $r=R$ .

Use Ampere's circuital law

For a long cylindrical current distribution, by symmetry the magnetic field is tangential and depends only on radial distance $r$ .

Thus,

\oint \vec B \cdot d\vec l = B(2\pi r)=\mu_0 I_{\text{enc}}

Now consider two regions.

Region 1: Inside the hollow cylinder ( $r<R$ )

Since all the current lies on the surface at $r=R$ , an Amperian loop of radius $r<R$ encloses no current:

I_{\text{enc}}=0

Hence,

B(2\pi r)=0 \implies B=0

So,

B=0 \quad \text{for } r<R

Region 2: Outside the cylinder ( $r>R$ )

Now the Amperian loop encloses the full current $I$ :

I_{\text{enc}}=I

Therefore,

B(2\pi r)=\mu_0 I

B=\frac{\mu_0 I}{2\pi r}

So,

B=\frac{\mu_0 I}{2\pi r} \quad \text{for } r>R

Behavior at $r=R$

Just inside the surface:

B_{\text{inside}}=0

Just outside the surface:

B_{\text{outside}}=\frac{\mu_0 I}{2\pi R}

So the graph has:

$B=0$ for all $r<R$
a sudden jump at $r=R$
then decreases as $1/r$ for $r>R$

Match with the correct graph

The correct representation must show:

zero field inside,
discontinuity at $r=R$ ,
inverse radial decay outside.

That corresponds to Option D.

Comparison with stored answer

Stored correct answer = D.

My derived answer = D.

So they agree.

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