A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cyl

Q: Why is the magnetic field inside the hollow cylinder zero? Doesn't the current on the surface create a field inside?

According to Ampere's Circuital Law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$. If you draw a circular Amperian loop inside the hollow region ($r < R$), the current enclosed by this loop is exactly zero. Due to the high symmetry of the cylinder, this implies that the magnetic field itself must be zero everywhere inside the hollow cavity.

Q: Does the magnetic field of the cylinder affect the magnetic field of the solenoid?

No. In classical electromagnetism (in linear media like vacuum), the Principle of Superposition holds perfectly. The magnetic field produced by the cylinder is completely independent of the magnetic field produced by the solenoid. You just calculate them separately and add them as vectors.

Q: In the region $R < r < 2R$, why isn't the field just along the common axis?

Because you are outside the hollow cylinder, it produces a magnetic field that wraps around it in circles (tangential direction). At the same time, the solenoid produces a field along the axis. Since both fields exist simultaneously and are perpendicular to each other, their vector sum points in a slanted direction, creating a helical field pattern.

Q: Why is the magnetic field outside the solenoid zero? Shouldn't there be some field leaking out?

For an *ideal, infinitely long* solenoid, the magnetic field outside is exactly zero. This is a standard theoretical result derived from Ampere's Law. While a real, finite solenoid does have a weak 'fringe' field outside, JEE problems always assume ideal infinite solenoids unless stated otherwise.

Q: How do we know the directions of the fields?

Use the Right-Hand Grip Rule. For the cylinder, point your thumb in the direction of the current along the axis; your curling fingers show the tangential magnetic field. For the solenoid, curl your fingers in the direction of the current wrapping around the tube; your thumb points in the direction of the axial magnetic field inside.

Question

A steady current $I$ flows along an infinitely long hollow cylindrical conductor of radius $R$. This cylinder is placed coaxially inside an infinite solenoid of radius $2R$. The solenoid has $n$ turns per unit length and carries a steady current $I$. Consider a point $P$ at a distance $r$ from the common axis. The correct statement(s) is (are)

Options:

(A) In the region $0 < r < R$, the magnetic field is non-zero

(B) In the region $R < r < 2R$, the magnetic field is along the common axis

(C) In the region $R < r < 2R$, the magnetic field is tangential to the circle of radius $r$, centered on the axis

(D) In the region $r > 2R$, the magnetic field is non-zero

Sigma Percentile · Accepted Answer

This problem is a beautiful exercise in the **Principle of Superposition** applied to magnetic fields. We are given two distinct, classic current distributions: an infinitely long hollow cylindrical conductor and an infinite solenoid, placed coaxially. To find the net magnetic field at any point, we simply need to evaluate the magnetic field produced by each source individually and then take their vector sum. ### Analyzing the Individual Sources Let's first recall the magnetic field profiles for our two sources using Ampere's Circuital Law. **1. The Hollow Cylindrical Conductor (Radius $R$)** For a hollow cylinder carrying a steady current $I$ along its length, the magnetic field inside the hollow region is zero because an Amperian loop inside encloses no current. Outside the cylinder, it behaves like a solid wire, producing a tangential magnetic field. - Inside ($r < R$): $B_{cyl} = 0$ - Outside ($r > R$): $B_{cyl} = \frac{\mu_0 I}{2\pi r}$ (Directed tangentially, i.e., in the $\hat{\phi}$ direction) **2. The Infinite Solenoid (Radius $2R$)** An ideal infinite solenoid carrying a current $I$ with $n$ turns per unit length produces a uniform magnetic field strictly confined to its interior, directed along its central axis. The field outside is zero. - Inside ($r < 2R$): $B_{sol} = \mu_0 n I$ (Directed axially, i.e., in the $\hat{k}$ direction) - Outside ($r > 2R$): $B_{sol} = 0$ ### Evaluating the Regions Now, we superimpose these fields in the three distinct regions defined by the geometry. **Region 1: The Innermost Core ($0 < r < R$)** Here, we are inside the hollow cylinder, so $B_{cyl} = 0$. However, we are still well inside the solenoid, meaning $B_{sol} = \mu_0 n I$. $$\vec{B}_{net} = \vec{B}_{cyl} + \vec{B}_{sol} = 0 + \mu_0 n I \hat{k} eq 0$$ The net magnetic field is non-zero and purely axial. This makes **Option (a)** correct. **Region 2: The Annular Gap ($R < r < 2R$)** In this intermediate region, we have stepped outside the hollow cylinder, so it now contributes a tangential magnetic field $B_{cyl} = \frac{\mu_0 I}{2\pi r}$. We are still inside the solenoid, so the axial field $B_{sol} = \mu_0 n I$ is also present. $$\vec{B}_{net} = \frac{\mu_0 I}{2\pi r} \hat{\phi} + \mu_0 n I \hat{k}$$ Because the net field is the vector sum of a tangential component and an axial component, the resultant magnetic field lines will trace out a **helical** path. It is neither purely axial nor purely tangential. This makes **Options (b) and (c)** incorrect. **Region 3: The Exterior ($r > 2R$)** Finally, we move completely outside the solenoid. The solenoid's contribution drops to zero ($B_{sol} = 0$). However, we are still outside the infinite cylinder, which continues to exert its tangential field. $$\vec{B}_{net} = \frac{\mu_0 I}{2\pi r} \hat{\phi} + 0 eq 0$$ The net magnetic field is non-zero and purely tangential. This makes **Option (d)** correct. ### Conclusion By systematically applying Ampere's Law and the superposition principle, we can confidently conclude that the magnetic field is non-zero in both the innermost and outermost regions. The correct statements are indeed (a) and (d).

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Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$M a g n e t i c F i e l d o f C y l in d er$

$M a g n e t i c F i e l d o f S o l e n o i d$

$R e g i o n 1 : 0 < r < R$

$R e g i o n 2 : R < r < 2 R$

$R e g i o n 3 : r > 2 R$

$F ina l C o n c l u s i o n$

The Sigma Insight: Ampere's Circuital Law

Analyzing the Individual Sources

Evaluating the Regions

Conclusion By systematically applying Ampere's Law and the superposition principle, we can confidently conclude that the magnetic field is non-zero in both the innermost and outermost regions

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field of Cylinder

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field of Solenoid

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Region 1: 0<r<R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Region 2: R<r<2R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Region 3: r>2R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion