Step-by-Step Guide: Magnetic Field Formula and Derivation Inside a Cylinder
The magnetic field due to an infinite current-carrying wire situated inside a cylinder is a fundamental topic in magnetostatics, relevant for JEE Main and other competitive exams. Understanding its mathematical form, variation, and derivation using Ampère’s Law is essential for accurate problem-solving in the context of cylindrical conductors.
Magnetic Field Due to an Infinite Straight Wire and Cylinder: Basic Concepts
A long straight wire carrying current produces a magnetic field in the space surrounding it. When such a wire has a cylindrical geometry, the field distribution depends on the radial position relative to the axis and the type of cylinder—solid or hollow. The analysis primarily uses Ampère’s Law and the concept of current distribution.
The study of this phenomenon forms an integral part of the syllabus under topics such as the Magnetic Effects Of Current, which details the origin and characteristics of the magnetic field from moving charges.
Ampère’s Law and its Application to Cylindrical Conductors
Ampère’s Law relates the integrated magnetic field around a closed loop to the current passing through the loop. For cylindrical symmetry, the law is applied along concentric circular paths centered on the axis of the current-carrying conductor. The quantitative relationship is given by:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$
Here, $\mu_0$ is the permeability of free space and $I_{\text{enc}}$ is the current enclosed by the Amperian loop. For infinite cylinders, this approach provides the field at any point in space.
Fundamentals involving straight wires can be reviewed at Magnetic Field Due To Straight Wire.
Magnetic Field Inside and Outside an Infinite Solid Cylinder
Consider a solid cylinder of radius $R$ carrying a uniform current $I$. The magnetic field at a distance $r$ from its axis depends on whether $r
| Region | Magnetic Field (B) |
|---|---|
| Inside cylinder ($r < R$) | $B = \dfrac{\mu_0 I r}{2\pi R^2}$ |
| On surface ($r = R$) | $B = \dfrac{\mu_0 I}{2\pi R}$ |
| Outside cylinder ($r > R$) | $B = \dfrac{\mu_0 I}{2\pi r}$ |
Within the solid cylinder, the magnetic field grows linearly with distance from the axis. Outside, it decreases inversely with distance. The maximum field lies at the surface ($r = R$).
This principle is also related to Magnetic Flux and geometry-specific cases.
Magnetic Field in Hollow Cylindrical (Cylindrical Shell) Conductors
A hollow cylindrical shell with inner radius $a$ and outer radius $b$ exhibits different field behaviors in distinct radial segments. The current is assumed to be uniformly distributed across the shell thickness.
For $r < a$ (inside the hollow): $B = 0$. For $a < r < b$ (within the shell): $B$ depends on the current enclosed and varies as $B \propto (r^2 - a^2)$. For $r > b$: $B = \dfrac{\mu_0 I}{2\pi r}$.
The right-hand rule determines the field direction, which is tangent to the concentric circles around the axis.
Derivation of Magnetic Field: Stepwise Approach Using Ampère’s Law
To derive the field at any distance $r$ within a solid cylinder ($r < R$), one considers the uniform current density $J = \dfrac{I}{\pi R^2}$. The current within radius $r$ is $I_{\text{enc}} = J \pi r^2 = I \dfrac{r^2}{R^2}$. Applying Ampère’s Law along a circular path of radius $r$ yields:
$B \cdot 2\pi r = \mu_0 I_{\text{enc}}$
$B \cdot 2\pi r = \mu_0 I \dfrac{r^2}{R^2}$
$B = \dfrac{\mu_0 I r}{2\pi R^2}$
For points outside ($r > R$), the enclosed current is $I$, so $B = \dfrac{\mu_0 I}{2\pi r}$.
This procedure involves key steps also found in the Biot-Savart Law for magnetic field determination in other symmetries.
Variation and Graphical Representation of Magnetic Field in Cylinders
For a solid cylinder, the graph of magnetic field $B$ versus radius $r$ increases linearly from $r=0$ to $r=R$ and then declines as $1/r$ beyond $r=R$. In hollow cylinders, $B=0$ in the inner cavity, increases within the shell, and decreases as $1/r$ outside.
The direction of the magnetic field remains tangent to circles centered on the axis and is given by the right-hand grip rule, matching the current direction.
Variation patterns can be studied in the context of Self-Inductance where different conductor morphologies are analyzed.
Comparison Table: Solid Cylinder vs Hollow Cylinder
| Aspect | Solid Cylinder |
|---|---|
| Field inside | Linear increase ($B \propto r$) |
| Field in cavity ($r < a$ for hollow) | Zero (no current enclosed) |
| Field outside | Inverse ($B \propto 1/r$) |
| Direction | Tangential (right-hand rule) |
Solved Example: Magnetic Field Calculation in a Solid Cylinder
A long solid cylinder of radius $2.0\, \mathrm{cm}$ carries a uniform current of $8.0\, \mathrm{A}$. Find the magnetic field at:
- $r = 1.0\, \mathrm{cm}$ (inside)
- $r = 2.0\, \mathrm{cm}$ (surface)
- $r = 3.0\, \mathrm{cm}$ (outside)
For $r = 1.0\,\mathrm{cm}$: $B = \dfrac{\mu_0 I r}{2\pi R^2}$. Substitute $r=0.01\,\mathrm{m}$, $R=0.02\,\mathrm{m}$, $I=8.0\,\mathrm{A}$, $\mu_0=4\pi\times10^{-7}\, \mathrm{T m/A}$. $B = \dfrac{4\pi\times10^{-7} \times 8 \times 0.01}{2\pi \times (0.02)^2} = 8\times10^{-5}\, \mathrm{T}$.
For $r=2.0\,\mathrm{cm}$ (surface): $B = \dfrac{\mu_0 I}{2\pi R} = \dfrac{4\pi\times10^{-7} \times 8}{2\pi \times 0.02} = 8\times10^{-5}\, \mathrm{T}$.
For $r = 3.0\,\mathrm{cm}$ (outside): $B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{4\pi\times10^{-7} \times 8}{2\pi \times 0.03}= 5.33\times10^{-5}\, \mathrm{T}$.
Key Points, Formula Summary, and Practical Tips
- Always confirm region (inside, shell, or outside) before applying formulas
- Use $B = \dfrac{\mu_0 I r}{2\pi R^2}$ for $r < R$ in solid cylinders
- Use $B = \dfrac{\mu_0 I}{2\pi r}$ for $r \geq R$ or $r \geq b$
- In hollow regions ($r < a$), $B = 0$
- Direction follows right-hand rule
- Stay consistent with SI units for calculations
FAQs on Understanding Magnetic Fields From Infinite Wires in Cylinders
1. What is the magnetic field due to an infinite straight wire in a cylinder?
The magnetic field due to an infinite straight wire placed inside a cylinder depends on the distance from the wire and the position inside or outside the cylinder. For a point at a distance r from the wire (where r is the radial distance), the field inside the cylinder is given by:
- B = (μ₀ I) / (2π r) (Ampere's Law), where μ₀ is the permeability of free space and I is the current.
- The direction of the field follows the right-hand thumb rule.
- The field forms concentric circles around the wire.
2. How do you calculate the magnetic field inside a cylinder containing an infinite wire?
To calculate the magnetic field inside a cylinder due to an infinite wire, use Ampere’s circuital law:
- Draw an Amperian loop of radius r centered on the wire.
- Apply: B × 2π r = μ₀ I
Thus, the field at distance r is B = (μ₀ I) / (2π r).
3. What is the direction of magnetic field around an infinite current-carrying wire inside a cylinder?
The direction of the magnetic field is tangent to the circle centered on the wire and determined by the right-hand rule:
- Grip the wire so your thumb points in the direction of current (I).
- Your fingers curl in the direction of the magnetic field lines (concentric circles around the wire).
4. How does the magnetic field due to an infinite wire vary with distance inside a cylinder?
The magnetic field strength due to an infinite wire in a cylinder decreases as the distance from the wire increases:
- B ∝ 1/r, where ‘r’ is the distance from the wire.
- Closer to the wire, the magnetic field is stronger.
- As you move away, the field decreases proportionally to the inverse of distance.
5. What is the formula for the magnetic field inside a long cylindrical conductor carrying current?
Inside a long cylindrical conductor of radius R carrying uniformly distributed current I:
- For a point at distance r < R, B = (μ₀ I r) / (2π R²)
- For r ≥ R, B = (μ₀ I) / (2π r)
This uses Ampere's Law and assumes uniform distribution of current inside the conductor.
6. Why does the magnetic field inside a hollow cylinder due to an infinite wire differ from that inside a solid cylinder?
For a hollow cylinder, if the current only flows in a central wire, the magnetic field does not depend on the material of the cylinder but only on the distance from the wire:
- Field lines are still concentric circles.
- If the cylinder does not carry current, the field remains B = (μ₀ I) / (2π r).
- In a solid cylinder with distributed current, the field within varies linearly with r.
7. State Ampere's circuital law and its application in finding magnetic field due to an infinite wire in a cylinder.
Ampere’s circuital law states that the line integral of magnetic field B around a closed loop equals μ₀ times the total current enclosed:
- ∮B⋅dl = μ₀ Iₑₙ𝚌
- For an infinite wire, choose an Amperian loop centered on the wire.
- This gives: B × 2π r = μ₀ I leading to B = (μ₀ I) / (2π r).
8. What is the significance of the right-hand rule in determining the magnetic field direction around an infinite wire?
The right-hand rule is essential for finding the circular direction of the magnetic field around an infinite current-carrying wire:
- Point your right thumb along the direction of current.
- Curl your fingers: they show the direction of the magnetic field lines.
- This helps remember that field lines form concentric circles around the wire, following the current's direction.
9. What are the real-life applications of the magnetic field produced by an infinite wire in a cylinder?
This physical principle is the basis for many electromagnetic devices:
- Electromagnets
- Transformers
- Inductors and solenoids
- Power transmission lines
Understanding magnetic fields inside cylinders helps in electrical engineering, especially in cable design and electromagnetic shielding.
10. Derive the expression for the magnetic field at a point inside a solid cylindrical conductor carrying current.
To derive the magnetic field inside a solid cylindrical conductor (radius R), carrying current I distributed uniformly:
- Take an Amperian loop at distance r < R from center.
- Current enclosed, I' = I (π r² / π R²) = I r² / R²
- Apply Ampere’s Law: B × 2π r = μ₀ I r² / R²
- Solve for B: B = (μ₀ I r) / (2π R²)
