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Electricity and Magnetism
Magnetic Field on the Axis of a Circular Coil
The magnetic field at any axial point of a current-carrying circular coil is directed along its axis and is derived using the Biot-Savart law. It forms the foundation for calculating magnetic fields of complex winding geometries in electrical machines.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Calculation of magnetic field at the center and axial points of a circular coil, variation of field with distance, and application of the Right-Hand Rule.
PGT extension
Derivation of the axial field formula using Biot-Savart law, spatial second derivative of the field, Helmholtz coil configuration for uniform magnetic field, and integration over multiple turns.
Prerequisites
Biot-Savart Law and its vector form
Integration techniques for trigonometric functions
Right-hand thumb and palm rules for magnetic fields
Biot-Savart Law Application to a Circular Loop
Consider a circular coil of radius R carrying current I with N turns, lying in the y-z plane with its center at the origin.
An elemental current length Idl is perpendicular to the position vector r connecting the element to an axial point P at distance x.
By the Biot-Savart law, the magnitude of the magnetic field dB due to Idl is independent of the azimuthal angle around the loop by symmetry.
Symmetry and Vector Components
The magnetic field vector dB makes an angle with the axis of the coil due to the cross product in the Biot-Savart law.
Perpendicular components (dB_y) cancel out diametrically opposite elements, leaving only the axial components (dB_x) to contribute to the net magnetic field.
The net magnetic field is directed entirely along the axis of the coil (the x-axis).
Special Case: Center of the Coil
At the center of the circular coil, the axial distance x becomes zero.
The general axial formula reduces to the maximum field value at the center: $B = \mu_0 N I / (2R)$.
The field lines are straight and parallel near the center, representing a uniform magnetic field over a small region.
Inflection Points and Helmholtz Coils
The derivative of the axial magnetic field vanishes at $x = \pm R/2$, which correspond to points of inflection where the field gradient is maximum.
When two identical coaxial coils are separated by a distance equal to their radius R, their combined field has a region of high uniformity between them, known as a Helmholtz coil setup.
In a Helmholtz coil, the first and second spatial derivatives of the magnetic field vanish at the midpoint, ensuring uniform magnetic field distribution.
Derivations
Derivation of Magnetic Field on the Axis of a Circular Coil
Conditions: Coil of radius R, N turns, current I, point P at axial distance x.
Write the Biot-Savart law for a current element Idl: d$B = (\mu_0 / 4\pi) * (I |dl \times r|) / r^3$.
Recognize that angle between dl and r is 90 degrees, so |dl \times r| = dl * r.
Resolve dB into axial (dB cos\theta) and radial (dB sin\theta) components, where cos$\theta = R / \sqrt{R^2 + x^2}$.
Integrate dB cos\theta around the loop: $B = \int dB (R / \sqrt{R^2 + x^2}) = (\mu_0 I R / 4\pi (R^2 + x^2)^{3/2}) \int dl$.
Since \int dl = 2\pi R for a single turn and multiplying by N turns, obtain the final result.
Result: $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$
Derivation of Position of Inflection Points (d2B/dx2 = 0)
Conditions: Single circular coil of radius R, finding points where curvature of B(x) changes sign.
Start with B(x) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}.
Find the first derivative dB/d$x = -\frac{3 \mu_0 I R^2 x}{2(R^2 + x^2)^{5/2}}$.
Apply the quotient rule to find the second derivative $d^2B/dx^2 = -\frac{3 \mu_0 I R^2}{2} * \frac{(R^2 + x^2)^{5/2} - x * (5/2)(R^2 + x^2)^{3/2}(2x)}{(R^2 + x^2)^5}$.
Simplify the numerator by factoring out (R^2 + x^2)^{3/2}: numerator proportional to (R^2 + x^2) - 5$x^2 = R^2 - 4x^2$.
Set $d^2B/dx^2 = 0$ to solve for x: R^2 - 4$x^2 = 0$.
Result: x = \pm \frac{R}{2}
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Axial Magnetic Field of Circular Coil
$$B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$$
B = Magnetic field, \m$u_0 = Permeability of free space$, N = Number of turns, I = Current, R = Radius, x = Axial distance Tesla (T)
Valid for a tightly wound circular coil of radius R
Magnetic Field at the Center
$$B_{center} = \frac{\mu_0 N I}{2R}$$
$B_{center} = Field at x=0$, N = Number of turns, I = Current, R = Radius Tesla (T)
At the geometric center of the coil
Magnetic Field at Large Distances (x >> R)
$$B = \frac{\mu_0 N I R^2}{2x^3} = \frac{\mu_0 2 M}{4\pi x^3}$$
M = NIA = Magnetic dipole moment of the coil, x = Distance Tesla (T)
Valid when axial distance x is much greater than radius R
Helmholtz Coil Magnetic Field
$$B_{total} = \left( \frac{4}{5} \right)^{3/2} \frac{\mu_0 N I}{R}$$
$B_{total} = Combined field at the midpoint$, separation d = R Tesla (T)
Two identical coaxial coils separated by distance equal to their radius
First Derivative of Axial Field
$$\frac{dB}{dx} = -\frac{3 \mu_0 N I R^2 x}{2(R^2 + x^2)^{5/2}}$$
Rate of change of magnetic field along the axis T/m
Any axial point x
Permeability of Free Space
$$\mu_0 = 4\pi \times 10^{-7}$$
\m$u_0 = Magnetic constant$ T m A^{-1} or H/m
Vacuum or air medium
Magnetic Dipole Moment of Coil
$$M = N I A = N I \pi R^2$$
M = Magnetic moment, A = Area of cross section A m^2
Draw: A bell-shaped symmetrical curve centered at x = 0, with maximum value B_0 at origin, inflection points at $x = \pm R/2$ where curvature changes, and asymptotic approach to zero as x approaches infinity.
Interpret: Shows that the magnetic field is maximum at the center and drops off non-linearly with distance, falling rapidly when x exceeds R.
Exam use: Frequently asked in multiple-choice questions to identify inflection points ($x = R/2)$ and the general shape of the decay curve.
Helmholtz Coil Uniform Field Region
Draw: Two individual bell-shaped curves overlapping such that their inflection points coincide, resulting in a nearly flat-topped resultant magnetic field curve in the central region.
Interpret: Demonstrates how combining two coils at a separation equal to their radius produces a region of nearly constant magnetic field strength.
Exam use: Conceptual questions on achieving uniform magnetic fields in laboratories.
Magnetic Field Lines around a Circular Coil
Draw: Closed loops emerging through the center of the coil, looping around the conductor wire, densely packed at the center and spreading out further away.
Interpret: Visualizes the direction and density of the magnetic flux, confirming the magnetic dipole behavior of the current loop.
Exam use: Pattern identification and matching questions in TGT and PGT examinations.
Solved numericals
TGT core
A circular coil of 100 turns has a radius of 0.1 m and carries a current of 2 A. Calculate the magnetic field at its center.
Identify given values: N = 100, R = 0.1 m, I = 2 A, \m$u_0 = 4\pi \times 10^{-7} T m A^{-1}$.
Use formula for field at the center: $B = \mu_0 N I / (2R)$.
Two identical coaxial circular coils each of radius 0.5 m and 50 turns are placed 0.5 m apart. If a current of 1 A flows through both in the same direction, find the magnetic field at the midpoint between them.
Identify that the separation d equals the radius R (0.5 m), which satisfies the Helmholtz coil condition.
Use the Helmholtz coil formula: $B_{total} = (4/5)^{3/2} * (\mu_0 N I / R) * 2 (since there$ are two identical coils).
Assumes the coil wire is thin and the current is uniformly distributed across its cross-section.
Edge effects and finite thickness of wire insulation are neglected in standard academic derivations.
Formula assumes free space or non-magnetic medium (\mu_r \approx 1) unless explicitly modified by core permeability.
Common misconceptions
Common error: The magnetic field is maximum at points other than the center along the axis.
Correct: The magnetic field is always maximum at the center ($x = 0) of a circular current-carrying coil$.
The denominator (R^2 + x^2)^{3/2} is minimum when x = 0, making the field magnitude maximum at the center.
Common error: Radial components of the magnetic field do not exist anywhere on the axis.
Correct: Radial components cancel out strictly along the axis due to symmetry, but exist off-axis.
Diametrically opposite current elements produce radial components that cancel out only on the symmetrical central axis.
Quick revision
Axial magnetic field formula: $B = \mu_0 N I R^2 / 2(R^2 + x^2)^{3/2}$.
Axial Magnetic Field of Circular Coil: $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$
Magnetic Field at the Center: $B_{center} = \frac{\mu_0 N I}{2R}$
Must remember
The magnitude of the magnetic field at the axial distance $x$ from the center of a circular coil of radius $R$, $N$ turns, and current $I$ is given by $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$.
The magnetic field is maximum at the center ($x = 0$) of the circular coil, having the value $B_{center} = \frac{\mu_0 N I}{2R}$.
Perpendicular components of $dB$ cancel out due to diametrically opposite symmetry, leaving only the axial components along the $x$-axis.
Points of inflection (where the field gradient is maximum and the second spatial derivative vanishes) occur at axial distances $x = \pm R/2$.
In a Helmholtz coil setup, two identical coaxial coils are separated by a distance equal to their radius $R$, producing a highly uniform magnetic field region between them.
Glossary
$B$Magnetic field vector or its magnitude, measured in Tesla (T).
$\mu_0$Permeability of free space, equal to $4\pi \times 10^{-7} \text{ T m A}^{-1}$.
$M$Magnetic dipole moment of the coil, defined as $M = N I A$ or $N I \pi R^2$.
Helmholtz CoilsTwo identical coaxial circular coils separated by a distance $d = R$ used to generate a uniform magnetic field.
Point of InflectionAxial positions $x = \pm R/2$ where the first derivative of the magnetic field is extreme and its second derivative is zero.
Magnetic Field at Center vs. Large Axial Distance
Feature
At the Center ($x = 0$)
At Large Distance ($x \gg R$)
Formula
$B = \frac{\mu_0 N I}{2R}$
$B = \frac{\mu_0 2 M}{4\pi x^3}$
Distance Dependence
Independent of $x$
Inversely proportional to $x^3$
Field Uniformity
Uniform over a small region near the center
Resembles the field of a magnetic dipole
Practice Quiz
Choose an option to reveal the explanation.
1. What is the formula for the magnetic field at the center of a circular coil of $N$ turns, radius $R$, carrying current $I$?
Explanation: At the center of the circular coil, $x = 0$. Substituting $x = 0$ into the general axial formula $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$ yields $B = \frac{\mu_0 N I}{2R}$.
2. Along the axis of a circular current-carrying coil, which components of the elemental magnetic field $dB$ cancel out due to symmetry?
Explanation: Diametrically opposite current elements produce perpendicular (radial) components of $dB$ that point in opposite directions and cancel out, leaving only the axial components.
3. At what axial distance $x$ from the center do the points of inflection of the magnetic field occur?
Explanation: The derivative of the axial magnetic field vanishes at $x = \pm R/2$, which correspond to points of inflection where the field gradient is maximum.
4. What is the separation distance between two identical coaxial coils in a Helmholtz coil configuration?
Explanation: In a Helmholtz coil setup, two identical coaxial coils are separated by a distance equal to their radius $R$ ($d = R$) to create a region of high magnetic field uniformity.
5. How does the magnetic field $B$ vary with axial distance $x$ when $x$ is much greater than the radius $R$ ($x \gg R$)?
Explanation: When $x \gg R$, the term $R^2$ in the denominator $(R^2 + x^2)^{3/2}$ can be neglected, reducing the expression to $B = \frac{\mu_0 N I R^2}{2x^3}$, meaning $B \propto \frac{1}{x^3}$.
6. What is the magnetic dipole moment $M$ of a planar circular coil with $N$ turns, radius $R$, and current $I$?
Explanation: The magnetic dipole moment of a planar circular coil is given by $M = N I A$, where the area $A = \pi R^2$, thus $M = N I \pi R^2$.
7. What is the unit of the permeability of free space $\mu_0$?
Explanation: The SI unit of $\mu_0$ is Tesla meter per Ampere ($\text{T m A}^{-1}$) or Henry per meter ($\text{H/m}$).
8. Near the center of a circular current-carrying coil, the magnetic field lines are:
Explanation: Near the center of the circular coil, the field lines are straight and parallel, indicating a uniform magnetic field over a small region.
9. If the radius of a circular coil is doubled while keeping the current and number of turns constant, what happens to the magnetic field at its center?
Explanation: The center magnetic field is $B = \frac{\mu_0 N I}{2R}$. Since $B$ is inversely proportional to $R$, doubling $R$ will halve the magnetic field.
10. What is the ratio of the magnetic field at an axial distance $x = R$ to the magnetic field at the center $x = 0$ for a circular coil?
Explanation: Using the ratio formula $\frac{B(x)}{B(0)} = \left(1 + \frac{x^2}{R^2}\right)^{-3/2}$, substituting $x = R$ gives $(1 + 1)^{-3/2} = (2)^{-3/2} = \frac{1}{2\sqrt{2}}$.
11. Which of the following describes the first derivative of the axial magnetic field $\frac{dB}{dx}$?
Explanation: Differentiating $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$ with respect to $x$ using the chain rule gives $\frac{dB}{dx} = -\frac{3 \mu_0 N I R^2 x}{2(R^2 + x^2)^{5/2}}$.
12. What is the value of the combined magnetic field $B_{total}$ at the midpoint of a Helmholtz coil setup?
Explanation: The combined magnetic field at the midpoint of two identical coaxial coils separated by $d = R$ is given by $B_{total} = \left(\frac{4}{5}\right)^{3/2} \frac{\mu_0 N I}{R}$.
13. Why is the magnetic field maximum at the center ($x = 0$) of a circular coil?
Explanation: In the formula $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$, the numerator is constant for a given coil, and the denominator is minimized when $x = 0$, resulting in the maximum field magnitude.
14. What physical property is ensured when the first and second spatial derivatives of the magnetic field vanish at the midpoint of a Helmholtz coil?
Explanation: Vanishing first and second spatial derivatives at the midpoint ensure that the magnetic field is extremely flat and uniform across that region.
15. In which plane is the circular coil typically considered to lie when deriving its axial magnetic field using the Biot-Savart law in standard textbook problems?
Explanation: By convention in standard derivations, the circular coil is considered to lie in the y-z plane with its center at the origin, so its axis coincides with the x-axis.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A circular coil of radius $R$ and $N$ turns carries a current $I$. The magnetic field at a point on its axis at a distance $R$ from its center is proportional to:
Explanation: Substitute $x = R$ into the axial formula: $B = \frac{\mu_0 N I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 N I R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 N I R^2}{2(2^{3/2} R^3)} = \frac{\mu_0 N I}{4\sqrt{2} R}$. Thus, $B$ is proportional to $R^{-1}$.
2. The ratio of the magnetic field at the center of a circular coil to that at a distance equal to its radius along the axis is:
Explanation: The ratio of center field to axial field at $x = R$ is $\frac{B(0)}{B(R)} = \left(1 + \frac{R^2}{R^2}\right)^{3/2} = (2)^{3/2} = 2\sqrt{2}$.
3. Two identical circular coils of radius $R$ carry equal currents in the same direction and are placed coaxially at a separation $R$. The setup forms a:
Explanation: Two identical coaxial coils separated by a distance equal to their radius $R$ form a Helmholtz coil setup, which provides a uniform magnetic field.
4. At what distance along the axis of a circular coil of radius $R$ is the magnetic field intensity equal to $\frac{1}{8}$th of its value at the center?
Explanation: We need $\frac{B(x)}{B(0)} = \frac{1}{8} \implies \left(1 + \frac{x^2}{R^2}\right)^{-3/2} = \left(\frac{1}{2}\right)^3 \implies 1 + \frac{x^2}{R^2} = 4 \implies \frac{x^2}{R^2} = 3 \implies x = \sqrt{3}R$.
5. A circular loop of radius $r$ carrying a current $I$ has a magnetic moment $M$. If the loop is reshaped into a square of side $a$ carrying the same current $I$, what is the new magnetic moment?
Explanation: Perimeter remains equal: $2\pi r = 4a \implies a = \frac{\pi r}{2}$. Original moment $M = I(\pi r^2)$. New moment $M' = I(a^2) = I\left(\frac{\pi r}{2}\right)^2 = I \frac{\pi^2 r^2}{4} = \frac{\pi}{4} (I \pi r^2) = \frac{\pi}{4} M$.
6. The magnetic field at the center of a circular coil of radius $R$ is $B_1$. The magnetic field at a point on its axis at a distance $x = \sqrt{3}R$ is $B_2$. The ratio $\frac{B_1}{B_2}$ is:
Explanation: From the previous relation, when $x = \sqrt{3}R$, the ratio $\frac{B(x)}{B(0)} = \frac{1}{8}$. Thus, $\frac{B(0)}{B(x)} = \frac{B_1}{B_2} = 8$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the magnetic field at the center of a circular coil of 100 turns, radius $0.1\text{ m}$, carrying a current of $2\text{ A}$?
Explanation: Using $B = \frac{\mu_0 N I}{2R}$, we get $B = \frac{(4\pi \times 10^{-7}) \times 100 \times 2}{2 \times 0.1} = \frac{8\pi \times 10^{-5}}{0.2} = 4\pi \times 10^{-4}\text{ T}$.
2. The second derivative of the axial magnetic field of a circular coil becomes zero at:
Explanation: Points of inflection where the second spatial derivative of the magnetic field vanishes occur at $x = \pm R/2$.
3. If two identical coils of radius $R$ are set up as a Helmholtz coil with current $I$ and $N$ turns, what is the expression for the total magnetic field at the midpoint?
Explanation: Since $\left(\frac{4}{5}\right)^{3/2} = \frac{8}{5\sqrt{5}}$, the Helmholtz formula $B_{total} = \left(\frac{4}{5}\right)^{3/2} \frac{\mu_0 N I}{R}$ can also be written as $\frac{8\mu_0 N I}{5\sqrt{5} R}$.
4. At what axial distance does the magnetic field fall to $\frac{1}{2\sqrt{2}}$ of its maximum value at the center?
Explanation: The field is $\frac{1}{2\sqrt{2}}$ (or $\frac{1}{2^{3/2}}$) of the center value when $x = R$, because $\left(1 + \frac{R^2}{R^2}\right)^{-3/2} = (2)^{-3/2} = \frac{1}{2\sqrt{2}}$.
5. What is the magnetic dipole moment of a circular coil of radius $0.2\text{ m}$ having $50$ turns and carrying a current of $5\text{ A}$?
Explanation: $M = N I \pi R^2 = 50 \times 5 \times \pi \times (0.2)^2 = 250 \times \pi \times 0.04 = 10\pi \text{ A m}^2$? Wait: $250 \times 0.04 = 10$? Let's check: $50 \times 5 = 250$. $250 \times 0.04 = 10$. So $10\pi$. Wait, option 3 is $10\pi$. Let's check correct_index for $10\pi$. Option 2 is $5\pi$, Option 3 is $10\pi$. Index 2 is correct.
6. Which of the following statements is correct regarding the magnetic field produced by a circular current-carrying loop?
Explanation: Due to symmetry, perpendicular components cancel out, leaving the net magnetic field directed entirely along the axis of the coil.
7. What happens to the rate of change of the magnetic field $\frac{dB}{dx}$ at the center ($x = 0$) of a circular coil?
Explanation: From $\frac{dB}{dx} = -\frac{3 \mu_0 N I R^2 x}{2(R^2 + x^2)^{5/2}}$, substituting $x = 0$ gives $\frac{dB}{dx} = 0$, confirming a peak (maximum) in the magnetic field curve at the center.
8. For a circular coil of radius $R$, at large distances where $x \gg R$, the magnetic field varies inversely as:
Explanation: At large distances ($x \gg R$), $B = \frac{\mu_0 N I R^2}{2x^3}$, showing an inverse cube dependence ($x^3$).
9. If the number of turns in a circular coil is doubled and its radius is halved while keeping the current constant, how does the magnetic field at the center change?
Explanation: $B_{center} = \frac{\mu_0 N I}{2R}$. If $N \to 2N$ and $R \to R/2$, then $B' = \frac{\mu_0 (2N) I}{2(R/2)} = 4 \left(\frac{\mu_0 N I}{2R}\right) = 4 B_{center}$? Wait: denominator is $2(R/2) = R$, so $2N / 1 = 2$. Thus it becomes 2 times? Wait: $N$ is doubled ($2N$), $R$ is halved ($R/2$). $2N / (R/2) = 4N/R$. Original was $N/R$. So it becomes 4 times. Let's check options: Option 2 is 2 times, Option 3 is 4 times. Correct index is 2.
10. In a Helmholtz coil setup, why is the second derivative of the magnetic field set to zero at the midpoint?
Explanation: Setting the first and second derivatives to zero ensures that the curvature of the magnetic field profile vanishes at the midpoint, creating a region of optimal spatial uniformity.