Step-by-Step Derivation of the Magnetic Field Formula Using Biot–Savart Law
Magnetic Field On The Axis Of A Circular Current Loop is a classic topic in JEE Main Physics, connecting electricity and magnetism using fundamental principles. It explores how a steady current flowing through a circular wire generates a magnetic field at various points along the central axis perpendicular to the loop's plane. This principle is not only a favorite for derivation questions but also vital for applications like electromagnets and advanced electronics. Mastering the formula and its derivation helps with conceptual clarity and quick numerical problem solving in exams.
At its core, the topic uses the Biot–Savart law to deduce how the magnetic field varies with the loop's radius, the current, and the position on the axis. You may encounter this in circuit magnetism, electromagnetic field calculations, and experimental setups. Similar questions often appear in sections related to magnetic effects of current and magnetism and practicals.
Magnetic Field On The Axis Of A Circular Current Loop: Definition and Key Formula
The magnetic field at a specific point on the axis of a circular current loop is the net field produced by all elemental current segments. According to the Biot–Savart law, the field depends on the current (I), loop radius (R), and the distance along the axis from the center (x).
| Formula | Meaning of Terms |
|---|---|
| B = (μ₀IR²) / 2(R² + x²)3/2 | B = field at distance x on axis; μ₀ = permeability of free space; I = current; R = loop radius; x = axis distance from center |
This result is critical for quick calculations in objective and numerical problems on electricity and magnetism. The same technique is adjusted for related configurations like solenoids, rings, and coils.
Step-by-Step Derivation Using Biot–Savart Law
You must know or recall each algebraic step for JEE-style derivations. Here’s a concise, exam-ready sequence for magnetic field on the axis of a circular loop using the Biot–Savart law:
- Consider a circular loop, radius R, lying in the xy-plane, with center at O and current I circulating anticlockwise.
- Set the point P at distance x from the center on the axis perpendicular to the loop (z-axis).
- Take a small current element Id𝐥 on the loop; the field it produces at P, from the Biot–Savart law, is:
dB = (μ₀/4π) * (Id𝐥 × r̂) / r² - Because of symmetry, only the axial (z) components from each element add up. Radial components cancel.
- Net field at P (only z-component):
B = (μ₀I R²) / 2(R² + x²)3/2
Memorise these steps, focusing on symmetry and vector addition. Right-hand thumb rule helps you get the field’s direction along the axis. Always conclude with the simplified formula for calculations.
Applications and Visualisation of Magnetic Field on Axis of Circular Loop
To build intuition, visualise how field lines inside the loop resemble those of a bar magnet near the center, pointing along the axis. Including multiple loops amplifies the net field proportionally, useful in practical devices.
- Current loops as magnetic dipoles behave like small magnets.
- Used in electromagnet design and MRI scanner magnets.
- Forms the basis of moving coil galvanometers and speakers.
- Essential for coil-based experiment setups in electromagnetic induction.
For 3D understanding, imagine a nested set of field lines forming concentric circles in the plane and stretching out as straight lines along the symmetry axis. At the center (x = 0), the formula reduces to B = (μ₀I)/(2R) for a single loop.
Numerical Examples: JEE-Main Style Problems
Mastering the direct application of the magnetic field on the axis of a circular current loop formula is essential for exams. Here’s a standard example.
Example: Find the magnetic field at a point 10 cm from the center on the axis of a circular loop of radius 20 cm, carrying 5 A current. Use μ₀ = 4π × 10⁻⁷ T·m/A.
- Given: I = 5 A, R = 0.2 m, x = 0.1 m, μ₀ = 4π × 10⁻⁷.
- Formula: B = (μ₀ I R²)/2(R² + x²)^(3/2)
- Calculate denominator: R² + x² = 0.04 + 0.01 = 0.05
(0.05)^(3/2) = 0.05 × √0.05 ≈ 0.05 × 0.2236 ≈ 0.01118 - Numerator: μ₀ × I × R² = 4π × 10⁻⁷ × 5 × 0.04 ≈ 2.513 × 10⁻⁷
- B = (2.513 × 10⁻⁷) / (2 × 0.01118) ≈ 1.124 × 10⁻⁵ T
So, the magnetic field at 10 cm on the axis is 1.12 × 10⁻⁵ T.
Try similar problems with straight wires or coils to see the contrast.
Common Mistakes and Key Reminders When Using the Formula
- Mixing up field at the center (B = (μ₀I)/(2R)) vs. on the axis.
- Forgetting to cube the denominator: (R² + x²)3/2, not just square root.
- Using wrong direction: always check with the right-hand thumb rule.
- Missing units: field should be in Tesla (T) for SI.
- Leaving out factors for multiple turns: multiply by number of turns N if present.
Avoid these traps, especially under exam pressure. Link concepts to related principles like Gauss law or Faraday’s law when integrating advanced problems.
Quick Revision: Flash Notes on Magnetic Field On The Axis Of A Circular Current Loop
- Magnetic field at distance x on axis: B = (μ₀IR²)/2(R² + x²)^(3/2).
- Derived with Biot–Savart law, only axial components add.
- Field direction: Use right-hand thumb rule.
- For N turns, field is multiplied by N.
- Center value (x = 0): B = (μ₀I)/(2R) (important comparison).
- Application: Electromagnets, MRI, moving coil devices, induction coils.
For in-depth revision, access Vedantu’s magnetic effects of current and magnetism revision notes. Practise regularly and compare with solenoid field configurations for best JEE Main preparation.
This topic’s derivation and applications are set firmly in the JEE Main syllabus. Mastering the magnetic field on the axis of a circular current loop will strengthen your base for more complex electromagnetism and experiment-based questions.
FAQs on Magnetic Field on the Axis of a Circular Current Loop
1. What is the magnetic field on the axis of a circular current loop?
The magnetic field on the axis of a circular current loop is the magnetic field generated at any point along the central axis passing perpendicularly through the center of the loop (but not necessarily at the center itself). This field is crucial in JEE and CBSE exams. The formula for the field at a distance x from the loop's center along the axis is:
B = (μ₀ I R²) / (2 (R² + x²)3/2)
where:
• μ₀ is the permeability of free space
• I is the current in the loop
• R is the radius of the loop
• x is the distance from the center along the axis
This field is used in applications requiring uniform magnetic fields and is a key concept in exam questions.
2. What is the formula for the magnetic field at a point on the axis of a circular loop?
The formula for the magnetic field at a point on the axis of a circular current loop is:
B = (μ₀ I R²) / (2 (R² + x²)3/2)
where:
• B = Magnetic field at distance x along axis
• μ₀ = Permeability of free space (4π × 10⁻⁷ H/m)
• I = Current through the loop
• R = Radius of the loop
• x = Distance from loop center along the axis
This formula should be memorized for direct application in physics and engineering problems.
3. How do you derive the magnetic field due to a current-carrying circular loop?
The derivation uses the Biot–Savart law to calculate the magnetic field at a point on the axis of a circular current loop.
Key Steps:
1. Consider an element of the loop with current Idl.
2. Apply the Biot–Savart Law to compute the elemental field dB.
3. Resolve each elemental field into components; only the axial components add up due to symmetry.
4. Integrate over the entire loop.
5. Arrive at
B = (μ₀ I R²)/(2 (R² + x²)3/2).
Always show steps and state the assumptions clearly in your exams.
4. What is the direction of the magnetic field on the axis of a current loop?
The direction of the magnetic field on the axis of a current loop follows the right-hand thumb rule:
• If the fingers of your right hand curl in the direction of current flow, the thumb points along the direction of the magnetic field on the axis.
• On one side of the loop, the field points away from the loop, and on the other, towards it, depending on the current direction.
Remember: The field direction is always along the loop's axis and is essential for accurate diagram labeling and solving vector questions.
5. How does the radius or distance from the loop affect the field strength?
The magnetic field strength on the axis of a circular loop depends on both the radius (R) of the loop and the distance (x) from its center:
• As distance x increases from the center, the field B decreases rapidly.
• The field is maximum at the center (x = 0): B = μ₀ I / (2R).
• A larger radius (R) increases the field at the center, but for points far away, the field drops off faster.
• Formula: B = (μ₀ I R²) / (2 (R² + x²)3/2)
Understanding this helps solve numerical problems and compare field strengths at different positions.
6. What are real-world applications of the magnetic field on the axis of a circular loop?
Magnetic fields generated by circular current loops are used in many real-world applications and experiments:
• Electromagnets and MRI machines
• Loudspeakers and microphones
• Particle accelerators
• Galvanometers and moving-coil instruments
• Experimental setups to create nearly uniform magnetic fields for physics research
This concept is central to electronics, medical imaging, and engineering devices.
7. Is the field on the axis the same as at the center of the loop?
No, the field at the center is a special case of the field on the axis.
• At the center (x = 0): B = μ₀ I / (2R) (for a single loop)
• At a point on the axis (x ≠ 0): B = (μ₀ I R²)/[2(R² + x²)3/2]
The field decreases as you move away from the center along the axis, and this comparison is often asked in exams.
8. What happens to the magnetic field if the loop has multiple turns (N turns)?
For a circular loop with N turns, the magnetic field on the axis is N times stronger than that of a single turn:
• Use: B = [μ₀ N I R²]/[2(R² + x²)3/2]
• The field increases linearly with the number of turns.
• This adjustment is essential when dealing with coils, solenoids, or electromagnets.
Always check N before using the formula in numerical questions.
9. Can you use Ampère’s law for this setup? Why or why not?
Ampère’s law cannot be directly used for calculating the magnetic field on the axis of a circular loop.
• Reason: The required symmetry (as in infinite straight wires or solenoids) is absent.
• The field varies in direction and magnitude around the loop.
• The Biot–Savart law is the correct approach because it allows calculation for each infinitesimal segment and sums the effect.
Remember: Ampère’s law works best for cases with high symmetry (straight wire, toroid, solenoid).
10. What are common calculation mistakes in JEE/Boards questions on this topic?
Students often make a few common mistakes when solving magnetic field problems for a circular current loop:
• Mixing up center and axis formulas
• Incorrectly identifying direction of field (use right-hand thumb rule!)
• Ignoring the number of turns (N)
• Forgetting units or plugging in incorrect formula variables
• Calculation errors with exponents in denominator
Tip: Always write the formula in full, label diagrams, and check units before finalizing your answer.
