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Electromagnetism · Magnetic Field of a Loop

Magnetism - Magnetic Field in a Current-Carrying Wire Loop

Run a current through a loop of wire and measure the magnetic field it creates at the centre. Change the number of turns and watch the field grow in proportion, then use B = μ₀nI/(2R) to work backward and recover the loop's radius — checking that it stays constant across every trial.

Theory — Magnetic Field of a Current Loop

Currents Make Magnetic Fields

A moving electric charge creates a magnetic field. When current flows around a loop of wire, the magnetic field from every part of the loop adds together and is strongest at the centre, pointing along the loop's axis. The direction is given by the right-hand rule: curl the fingers of the right hand in the direction of the current and the thumb points along the field.

Field at the Centre of a Loop

For a circular loop (or a tight coil of n turns) carrying current I, the magnetic field at the centre depends on the current, the number of turns, and the loop radius.

Magnetic Field at the Centre of a Loop B = μ₀ · n · I / (2R)

B = magnetic field at the centre (T)
μ₀ = 4π × 10⁻⁷ T·m/A (permeability of free space)
n = number of turns, I = current (A), R = loop radius (m)
More turns or more current → stronger field; larger radius → weaker field

How the Field Responds

Because B is proportional to n and to I, doubling either one doubles the field. Because B is inversely proportional to R, doubling the radius halves the field. The magnetic field is a vector — it has both a magnitude (measured in tesla) and a direction (along the axis of the loop).

Recovering the Radius

If you measure B for a known current and number of turns, you can rearrange the formula to solve for the loop radius. Since the loop itself does not change, the radius you calculate should come out the same for every trial — a good check that the relationship holds.

Solving for the Radius R = μ₀ · n · I / (2B)
R should be constant across all trials for the same loop

More turns

B ∝ n. Each extra turn adds its own field, so the centre field grows in proportion to n.

More current

B ∝ I. A bigger current means faster-moving charge and a stronger field.

Bigger loop

B ∝ 1/R. Spreading the current over a larger loop weakens the field at the centre.

QuantityRelationshipNotes
Centre fieldB = μ₀nI/(2R)μ₀ = 4π×10⁻⁷ T·m/A
TurnsB ∝ ndouble n → double B
CurrentB ∝ Idouble I → double B
RadiusB ∝ 1/Rdouble R → half B
Solve for radiusR = μ₀nI/(2B)constant across trials

Instructions — Running the Virtual Experiment

The Loop & Field Meter tab lets you set the current and number of turns and read the field at the centre; the Field vs Turns tab records B against n so you can confirm the proportionality and recover the radius. Record every reading in your lab report with screenshots.

Part A — Measure the Field (Loop & Field Meter tab)
1
Open Simulation → Loop & Field Meter. The current is set to I = 3 A. Choose a number of turns n and read the magnetic field B at the centre from the field meter.
2
Repeat for at least three different numbers of turns (for example n = 1, 2, 3, 5, 10). Record n and the measured B each time.
3
For each trial, calculate the loop radius from R = μ₀nI/(2B). Confirm the radius comes out the same each time, then average it.
Part B — Field vs Turns (Field vs Turns tab)
1
Open Field vs Turns. Record each (n, B) point, then Plot & fit. The points lie on a straight line through the origin (B ∝ n), and the slope is μ₀I/(2R).
2
Confirm the radius from the slope, and check the trends: doubling the turns doubles the field; a larger loop would weaken it.

Simulation — Current-Carrying Loop

Magnetism Virtual LabSet the turns and read the field at the centre
current loop
field meter (centre)
I = 3 A · loop radius fixed · field along the axis.

Loop settings

Field meter
Current I3.0 A
Turns n1
Field B (centre)0.0377 mT
R = μ₀nI/(2B)0.0500 m
B = μ₀nI/(2R). Change n and watch B scale with it.
Each point is (n, B) at I = 3 A. Slope = μ₀I/(2R).

Record turns (I = 3 A)

Fit result
Slope (μ₀I/2R)— T/turn
→ radius R— m
Average R (trials)— m
Turns nB (mT)R = μ₀nI/(2B) (m)
Click a number of turns to record it.

Team Questions

Question 1. A single loop (n = 1) of radius 0.0500 m carries I = 3.0 A. Find B at the centre using B = μ₀nI/(2R), with μ₀ = 4π×10⁻⁷. (Type in mT to 3 sig figs — e.g. 0.0377)
Question 2. For the same loop with n = 2 turns, what is B at the centre? (Type in mT — e.g. 0.0754)
Question 3. When the number of turns doubled from 1 to 2, by what factor did the magnetic field change? (Type the factor — e.g. 2)
Question 4. Measuring B = 0.0377 mT (3.77×10⁻⁵ T) for n = 1, I = 3 A, solve for the loop radius R = μ₀nI/(2B). (Type in m — e.g. 0.0500)
Question 5. Why should the radius you calculate be the same for every trial, even as n changes? (One key idea)
Question 6. If the loop radius were doubled (same n and I), what would happen to the field at the centre? (Answer: halved / doubled / unchanged)
Question 7 — Challenge. On a graph of B (y) versus n (x), what does the slope of the line equal, in terms of μ₀, I, and R? (Type the expression)

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission, and include labelled screenshots of each loop configuration and field-meter reading.

Magnetism - Magnetic Field in a Current-Carrying Wire Loop

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To measure the magnetic field at the centre of a current-carrying wire loop for several numbers of turns, to verify that the field is proportional to the number of turns, and to use B = μ₀nI/(2R) to determine the loop radius — confirming that the calculated radius is consistent across all trials.

Theory

A current-carrying loop produces a magnetic field along its axis, strongest at the centre, given by B = μ₀nI/(2R), where μ₀ = 4π×10⁻⁷ T·m/A. The field is proportional to the number of turns and the current, and inversely proportional to the radius. Rearranged, R = μ₀nI/(2B), so measuring B for a known n and I gives the radius.

B = μ₀nI/(2R) · R = μ₀nI/(2B)
μ₀ = 4π×10⁻⁷ T·m/A, I = 3.0 A

Sample Calculation — n = 2

B = μ₀nI/(2R) = (4π×10⁻⁷)(2)(3.0)/(2 × 0.0500) = 7.54×10⁻⁵ T = 0.0754 mT

R = μ₀nI/(2B) = (4π×10⁻⁷)(2)(3.0)/(2 × 7.54×10⁻⁵) = 0.0500 m

Results Table (I = 3.0 A)

TrialTurns nMeasured B (mT)Calculated R (m)
110.03770.0500
220.07540.0500
330.11310.0500
450.18850.0500
5100.37700.0500

Average radius = 0.0500 m. A plot of B against n was a straight line through the origin with slope μ₀I/(2R) = 3.77×10⁻⁵ T/turn.

Discussion

The magnetic field increased in direct proportion to the number of turns: doubling n from 1 to 2 doubled B from 0.0377 to 0.0754 mT, and the n = 10 field was ten times the single-turn value. Solving R = μ₀nI/(2B) gave 0.0500 m for every trial, as expected, because the physical loop did not change — only the number of turns did. The graph of B versus n was linear through the origin, confirming B ∝ n, with a slope equal to μ₀I/(2R).

Small differences between trials would come from the resolution of the field meter and from reading the values, rather than from any real change in the radius. This experiment demonstrates the direct link between electricity and magnetism: an electric current alone, with no magnets present, produces a measurable magnetic field.

Conclusion

The magnetic field at the centre of the loop followed B = μ₀nI/(2R): it was proportional to the number of turns, and the radius recovered from each measurement was a consistent 0.0500 m. The results confirm the fundamental relationship between current and magnetic field.

Practice Questions

Show all work and include units. Use μ₀ = 4π×10⁻⁷ T·m/A and B = μ₀nI/(2R).

Question 1
A single circular loop of radius 0.10 m carries a current of 5.0 A. Find the magnetic field at the centre.
Hint: B = μ₀nI/(2R) with n = 1.
Question 2
A coil of 50 turns and radius 0.040 m carries 2.0 A. What is the magnetic field at its centre?
Hint: n = 50; the turns multiply the single-loop field.
Question 3
A field meter at the centre of a 20-turn coil reads 1.26 mT when the current is 4.0 A. Determine the radius of the coil.
Hint: R = μ₀nI/(2B).
Question 4
Explain why a graph of magnetic field versus number of turns passes through the origin and is a straight line. What is the physical meaning of its slope?
Hint: B = (μ₀I/2R)·n, so B ∝ n; slope = μ₀I/(2R).
Question 5
By what factor does the centre field change if you triple the current and at the same time double the radius?
Hint: B ∝ I and B ∝ 1/R, so factor = 3 × (1/2).
Question 6 — Challenge
A 100-turn coil of radius 0.050 m is to produce a field of 5.0 mT at its centre. What current is required?
Hint: solve I = 2RB/(μ₀n).