JEE Challenger
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PhyMedAdv 2013 • Paper 2

Magnetic Field due to Coaxial Hollow Cylinder and Solenoid

Consider a constant current II flowing through an infinitely long, hollow cylindrical conductor with radius RR. This cylinder is positioned coaxially within an infinite solenoid having radius 2R2R. The solenoid contains nn turns per unit length and carries a constant current II. Let PP be a point located at a distance rr from their common central axis. Identify the correct statement(s):

Options

A

Within the region 0<r<R0 < r < R, the magnetic field vector is non-zero.

Correct
B

In the interval R<r<2RR < r < 2R, the direction of the magnetic field lies along the common axis.

C

In the interval R<r<2RR < r < 2R, the magnetic field vector is tangential to a circle of radius rr centered on the common axis.

D

For the region r>2Rr > 2R, the magnetic field is non-zero.

Correct

Topics & Concepts

Moving Charges and MagnetismAmpere's Law

Step-by-Step Solution

To determine the correct statements, we analyze the magnetic field produced by the coaxial system in three different regions using Ampere's Circuital Law and the standard results for solenoids.

Let the central axis of the system be aligned along the zz-axis. We use cylindrical coordinates (r,ϕ,z)(r, \phi, z), where z^\hat{z} is the axial unit vector and ϕ^\hat{\phi} is the azimuthal (tangential) unit vector.

1. Field due to the Hollow Cylindrical Conductor (Radius RR)

Using Ampere's Circuital Law for a circular loop of radius rr coaxial with the cylinder:

  • For r<Rr < R: Since no current is enclosed by the loop, B⃗cyl=0⃗\vec{B}_{\text{cyl}} = \vec{0}
  • For r>Rr > R: The total enclosed current is II, so ∮B⃗cyl⋅dl⃗=Bcyl(2πr)=μ0I  ⟹  B⃗cyl=μ0I2πrϕ^\oint \vec{B}_{\text{cyl}} \cdot d\vec{l} = B_{\text{cyl}} (2\pi r) = \mu_0 I \implies \vec{B}_{\text{cyl}} = \frac{\mu_0 I}{2\pi r} \hat{\phi}

2. Field due to the Infinite Solenoid (Radius 2R2R)

For an ideal infinite solenoid with nn turns per unit length carrying current II:

  • For r<2Rr < 2R: The field inside is uniform and directed along the central axis: B⃗sol=μ0nIz^\vec{B}_{\text{sol}} = \mu_0 n I \hat{z}
  • For r>2Rr > 2R: The field outside the ideal infinite solenoid is zero: B⃗sol=0⃗\vec{B}_{\text{sol}} = \vec{0}

3. Superposition of Fields in Different Regions

Region I: 0<r<R0 < r < R

B⃗total=B⃗cyl+B⃗sol=0⃗+μ0nIz^=μ0nIz^\vec{B}_{\text{total}} = \vec{B}_{\text{cyl}} + \vec{B}_{\text{sol}} = \vec{0} + \mu_0 n I \hat{z} = \mu_0 n I \hat{z}

  • The magnetic field is non-zero and points purely along the central axis.
  • Statement (A) is correct.

Region II: R<r<2RR < r < 2R

B⃗total=B⃗cyl+B⃗sol=μ0I2πrϕ^+μ0nIz^\vec{B}_{\text{total}} = \vec{B}_{\text{cyl}} + \vec{B}_{\text{sol}} = \frac{\mu_0 I}{2\pi r} \hat{\phi} + \mu_0 n I \hat{z}

  • The net field vector has both a tangential component (ϕ^\hat{\phi}) and an axial component (z^\hat{z}).
  • Therefore, the net magnetic field vector is neither purely along the axis nor purely tangential.
  • Statement (B) and Statement (C) are both incorrect.

Region III: r>2Rr > 2R

B⃗total=B⃗cyl+B⃗sol=μ0I2πrϕ^+0⃗=μ0I2πrϕ^\vec{B}_{\text{total}} = \vec{B}_{\text{cyl}} + \vec{B}_{\text{sol}} = \frac{\mu_0 I}{2\pi r} \hat{\phi} + \vec{0} = \frac{\mu_0 I}{2\pi r} \hat{\phi}

  • The net magnetic field is non-zero (it is purely tangential due to the inner hollow cylinder).
  • Statement (D) is correct.

Conclusion

The correct statements are A and D.

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