Physics · Magnetism

Torque on a current-carrying loop

Q: What does Torque on a current loop describe physically?

A current-carrying loop has magnetic dipole moment $\vec\mu=NIA\hat{\vec n}$. Its direction follows the right-hand rule for conventional current. In uniform external field, $\vec\tau=\vec\mu\times\vec B$ produces aligning torque. Opposite conductor forces can cancel as net force while their couple rotates the loop. The magnitude formula uses angle between normal and field, not plane and field. A fixed bearing additionally requires projection onto its allowed axis. Area refers to one turn; for nonidentical turns, add individual moment vectors instead of blindly multiplying by one common area. Magnitude alone cannot specify the direction of rotation.

Q: How do I check units and scale for Torque on a current loop?

Dipole moment has unit $\mathrm{A\,m^2}$. Multiplying by tesla gives $\mathrm{N\,m}$, the conventional torque unit. Although this dimension also matches energy, torque is not itself performed work. Work additionally requires rotation and the integral of signed axial torque over angle. Twenty square centimetres equals $0.00200\,\mathrm{m^2}$; a linear centimetre conversion would be wrong. As a scale check, torque magnitude in the uniform model cannot exceed $N|I|AB$. It vanishes when normal is parallel or antiparallel to the field. Unit checking alone detects neither using plane angle instead of normal angle nor selecting an incorrect bearing axis.

A planar current-carrying frame can experience torque in a uniform field despite zero net force, aligning its dipole axis.

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Formula

Calculate

τ = N ∣ I ∣ A B sin θ

τ in N·m; N dimensionless turn count; I in A; A in m² per turn; B in T; θ to the normal in rad or °.

Variables & units

Variables & units
Symbol	Meaning	Unit
$τ$	Torque magnitude	N·m
$N$	Number of turns	dimensionless
$I$	Nonnegative current magnitude; normal follows actual conventional circulation	A
$A$	Area of one turn	m²
$B$	Magnitude of uniform external magnetic flux density	T
$θ$	Angle between actual dipole vector or current-oriented normal and external field	rad or °

LaTeX

\tau=N|I|AB\sin\theta

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Fixed dipole: normalized orientation potential minus cosine rises from minus one at parallel to plus one at antiparallel alignment. Signed angular-coordinate torque is minus sine and acts towards decreasing angle between zero and one hundred eighty degrees. — Original diagram for fixed dipole magnitude in stationary uniform field, following OpenStax §11.5. θ is angle to current-oriented normal. τθ is signed; its magnitude is NIAB sinθ. Parallel alignment has minimum potential and is stable; antiparallel is maximum and unstable, although both torques vanish.

Current loop torque calculator

Find the magnetic moment, torque and orientation energy of a flat coil in a uniform magnetic field, with all turns of equal area.

Calculation model

Find

Number of turns N

Number of turns in the coil, a whole number, e.g. 100.

Current I

Current through the coil; its sign does not change the magnitude, e.g. 0.5 A.

Enclosed area of one turn A

Area of a single turn, not of all turns together, e.g. 20 cm² for 4 cm by 5 cm.

External magnetic-field magnitude B

Flux density of the uniform external magnetic field, e.g. 0.2 T between strong magnet poles.

Angle between normal and field θ

Angle between the coil normal and the field, not the coil plane; torque peaks at 90°, e.g. 30°.

Filled with example values. Type your own and the result updates straight away.

Result: Torque |τ|

|τ| = 10 mN·m

With N = 100, I = 0.5 A, A = 20 cm², B = 0.2 T and θ = 30 °, |τ| = 10 mN·m.

Working · rearrange for |τ|, substitute, evaluate

1. Rearrange for |τ|
$|\tau|=N|I|AB\sin\theta$
2. Substitute values (in SI units)
- A = 20 cm² = 0.002 m²
- θ = 30 ° = 0.523599 rad
$100\cdot 0.5\cdot 0.002\cdot 0.2\cdot\sin\left(0.523598775598\right)=0.01$
3. Result
|τ| = 10 mN·m

Quick tips

Check the angle: θ lies between the coil normal and the field. If the field lines run along the coil plane, θ = 90° and the torque is largest; if the plane is perpendicular to the field, it is zero.
Estimate in two steps: first μ = N · I · A, then τ = μ · B · sin θ. 100 turns, 0.5 A and 20 cm² give 0.1 A·m², so at most 20 mN·m in 0.2 T.

Assumptions and limits

N positive safe integer turns, each with the same enclosed area and orientation; rigid planar loop in a uniform external field. θ is between the actual dipole vector and B, not the loop plane. Signed current is used only through |I|; magnitudes determine neither rotation direction nor bearing torque. A=0 is only a formal zero-area limit.

Derivation & background

Using current-oriented normal, $\vec\mu=NIA\hat{\vec n}$ and $\vec\tau=\vec\mu\times\vec B$ . θ is measured to the normal, not the plane. A fixed dipole is stable parallel and unstable antiparallel; both torque magnitudes vanish. Continuous motor rotation needs additional current/field control and energy balance. A bearing uses the component along its axis.

Exam blueprint

Validity range

Identical planar current-carrying turns in uniform external field; θ is between current-oriented normal and field. The shortcut gives magnitude; a bearing needs axial component. Treat nonuniform fields and nonidentical turns separately.

Derivation steps

Opposite conductor forces form a couple; total separation of their lines of action gives area dependence.

1Specify identical planar turns and uniform external field.
2The right-hand rule for conventional current defines positive area normal.
3For the chosen rectangle, two opposite sides are transverse to field; their forces are equal and opposite.
4Each force magnitude is $|I|bB$ ; total lever-arm separation is $a\sin\theta$ .
5The couple gives $\tau=|I|abB\sin\theta$ ; with $A=ab$ , N turns add their contributions.
6Define $\vec\mu=NIA\hat{\vec n}$ ; generally $\vec\tau=\vec\mu\times\vec B$ .
7Check normal angle, zero net force and any required projection onto an allowed rotation axis.

Rearrangements

Turn count

N=\frac{\tau}{|I|AB\sin\theta}

θ lies between current-oriented normal and field; all denominator factors must be nonzero. Round integer minimum targets upwards.

Task variant

100 identical turns of area 0.010 m² each carry 0.20 A. Uniform external field is 0.50 T; the current-oriented normal is perpendicular to it. Find torque magnitude.

Approach: $\tau=N|I|AB\sin\theta$ ; at $\theta=90^\circ$ , $\tau=0.10\,\mathrm{N\,m}$ .

Common mistakes

Use the angle to the loop plane although the formula requires the angle to its oriented area normal.

Define the normal from current direction. On reversing current, do not also make the current magnitude negative.

Exam context

Loop orientation, maximum torque and stable alignment: check normal, current magnitude and field uniformity.

Formula cluster

Magnetism

Wire forces create a couple; magnetic flux and orientation energy connect loop orientation to torque.

M = F·r·sin(θ)

Torque

ΦB = BA cos θ

Magnetic flux through a surface

F = B|I|ℓ sinθ

Magnetic force on a current-carrying wire

Worked example

A coil with $N=100$ , $I=0.20\,\mathrm A$ , $A=0.010\,\mathrm{m^2}$ in $B=0.50\,\mathrm T$ at $\theta=90^\circ$ to the current-oriented normal experiences $\tau=0.10\,\mathrm{N\,m}$ .

Applications

Describe DC motors and magnetic dipoles quantitatively.

How do you calculate with Torque on a current loop?

Here is how to work through a typical Torque on a current loop (τ = N|I|AB sinθ) task step by step:

1
Task
100 identical turns of area 0.010 m² each carry 0.20 A. Uniform external field is 0.50 T; the current-oriented normal is perpendicular to it. Find torque magnitude.
Solution path
Approach: $\tau=N|I|AB\sin\theta$ ; at $\theta=90^\circ$ , $\tau=0.10\,\mathrm{N\,m}$ .

Understand and apply the equation

Opposite wire forces can cancel as net force while still rotating a loop. The decisive angle lies between its current-oriented surface normal and the field. We follow the force couple, calculate a finite rotation and distinguish an aligned dipole from a continuously running motor.

Which angle and single-turn area determine current-loop torque?

A wire can be pushed; a complete loop can rotate. In a uniform magnetic field, forces on opposite sides cancel overall. Because they act at different positions, they can still produce a turning effect called torque. The orientation of the loop area matters. Its normal is an imaginary arrow perpendicular to the area; current circulation selects which of the two directions is positive.[1]

\begin{aligned}\vec\mu&=NIA\hat{\vec n}\\\vec\tau&=\vec\mu\times\vec B\\|\vec\tau|&=N|I|AB\sin\theta\end{aligned}

Identical planar turns in uniform external field. Here $I\ge0$ is current magnitude and the normal follows the right-hand rule for actual conventional circulation. A signed current coordinate instead requires a fixed reference normal.[1]

The magnetic dipole axis is perpendicular to the loop plane. Viewed from its positive tip towards the loop, positive current is counterclockwise. Area is the enclosed area of one turn, not wire cross-section or the combined surface of all turns. Turn count adds equally oriented contributions. For differing areas or orientations, add individual moment vectors instead. Magnitude alone does not provide rotation direction; that requires the cross product.[1]

A force couple gives torque

\begin{aligned}F&=|I|bB\\d_\perp&=a\sin\theta\\\tau&=Fd_\perp=|I|abB\sin\theta\\A&=ab\end{aligned}

Rectangle with sides a and b; the b sides remain transverse to field. $d_\perp$ is the full perpendicular distance between their lines of action.[2][1]

Two opposite forces act at different sides. Use the full separation of their lines of action, without an additional factor of two. About the centre each force uses half that separation, and their moments add to the same result. The couple has zero net force, so its torque vector is independent of shifting the reference point. For a constrained bearing, however, only the component along the allowed rotation axis contributes; total vector magnitude is not automatically shaft torque. Magnetic force on a current-carrying wire supplies the individual contributions to the force couple.[1]

Why is torque zero at both parallel and antiparallel alignment?

Torque magnitude and normal angle

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Original model curve for the 25-turn rectangular coil: this is nonnegative magnitude rather than signed axial torque. The two zeros differ in stability.[1]

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\begin{aligned}U(\theta)&=-\mu B\cos\theta\\\tau_\theta&=-\frac{\mathrm dU}{\mathrm d\theta}=-\mu B\sin\theta\end{aligned}

$\mu=|\vec\mu|$ is the fixed dipole magnitude; θ is the angle between the actual dipole vector and the stationary uniform field. The second line is signed torque for coordinate θ, not its nonnegative magnitude.[1]

Parallel alignment has minimum potential and zero torque. A small displacement produces restoring torque. Antiparallel alignment also has zero torque but maximum potential: small perturbations lead away from it. Without damping, aligning torque does not mean the loop immediately settles parallel to the field; it can oscillate about stable alignment. Maximum torque occurs at ninety degrees to the normal, when the loop plane lies parallel to the field.[1]

Ampere square metres and shaft torque

\begin{aligned}[\mu]&=\mathrm{A\,m^2}\\{}[\tau]&=\mathrm{N\,m}\\N&=\frac{\tau}{|I|AB\sin\theta}\\\tau_{\mathrm{axis}}&=\vec\tau\cdot\hat{\vec e}_{\mathrm{axis}}\end{aligned}

Torque is expressed in N·m despite sharing energy’s dimension; turn count must be physically attainable as an integer.[1]

Square centimetres convert to square metres with factor 10⁻⁴. A round turn uses area from loop radius, not wire diameter. Rearranging magnitude requires nonzero current, field, area and sine. Zero torque without known orientation cannot uniquely determine current. To find work during rotation, integrate signed axial torque over angle in radians; an instantaneous magnitude times an arbitrary angle does not replace that integral. Torque about a specified axis must therefore be distinguished from its magnitude before work is determined.[1]

Determine the angle using a weight

Given: A supported coil has fixed dipole magnitude $\mu=0.0800\,\mathrm{A\,m^2}$ in uniform $B=0.300\,\mathrm T$ . A weight acts through an ideal lever with constant perpendicular arm $a=0.0400\,\mathrm m$ : $m=0.0300\,\mathrm{kg}$ , $g=9.81\,\mathrm{m\,s^{-2}}$ . Both torques act about the same axis; other torques vanish. Find an orientation in the first quadrant.
Find: Find the angle to the current-oriented normal for torque balance.

Approach: balance magnetic axial torque against the opposing weight torque. The first-quadrant condition selects one of two possible angles.
$\mu B\sin\theta=mga$
Substitution: first calculate the dimensionless sine, then the angle.
$\begin{aligned}\sin\theta&=\frac{(0.0300)(9.81)(0.0400)}{(0.0800)(0.300)}=0.4905\\\theta&=\arcsin(0.4905)\approx29.37^\circ\end{aligned}$

Result: The angle is approximately 29.4°, with torque magnitude 11.772 mN·m.

Check: Check: required torque is below $\mu B=24.0\,\mathrm{mN\,m}$ . The other angle, $150.63^\circ$ , lies outside the specified range. Stability in a real bearing requires the complete angular dependence of the load; equilibrium alone does not prove stability.

Example: check area and lever arms

A tilted rectangular coil

Given: 25 identical turns with $a=0.120\,\mathrm m$ and $b=0.0800\,\mathrm m$ ; current 0.300 A, uniform external field 0.400 T, angle to current-oriented normal 60°. The b sides are transverse to field.
Find: Find dipole magnitude and torque; check with the force couple.

Approach: first find single-turn area, then moment and cross-product magnitude.
$\begin{aligned}A&=ab\\\mu&=NIA\\\tau&=\mu B\sin\theta\end{aligned}$
Substitution: each turn has area 0.00960 m².
$\begin{aligned}\mu&=(25)(0.300)(0.00960)\,\mathrm{A\,m^2}=0.0720\,\mathrm{A\,m^2}\\\tau&=(0.0720)(0.400)\sin60^\circ\,\mathrm{N\,m}\\&\approx0.0249415\,\mathrm{N\,m}\end{aligned}$

Result: Dipole magnitude 0.0720 A·m²; torque magnitude about 24.9 mN·m.

Check: Each force of the couple has magnitude $NIbB=0.240\,\mathrm N$ . Separation is $a\sin60^\circ\approx0.103923\,\mathrm m$ ; their product gives the same torque. Maximum magnitude at ninety degrees would be 28.8 mN·m.

Example: work through a finite angle

Alignment without constant torque

Given: Fixed magnetic dipole magnitude 0.120 A·m² in uniform stationary 0.250 T field. θ decreases from 120° to 30°. Current and geometry are controlled constant; consider the effective mechanical dipole potential change.
Find: Calculate potential change and work by magnetic torque.

Approach: for a fixed dipole, work equals negative change of its orientation potential.
$\begin{aligned}\Delta U&=-\mu B(\cos\theta_f-\cos\theta_i)\\W_{\mathrm{mag}}&=-\Delta U\end{aligned}$
Substitution: retain angular dependence throughout rotation.
$\begin{aligned}\Delta U&=-(0.120)(0.250)(\cos30^\circ-\cos120^\circ)\,\mathrm J\\&\approx-0.0409808\,\mathrm J\\W_{\mathrm{mag}}&\approx+0.0409808\,\mathrm J\end{aligned}$

Result: Orientation potential decreases by 40.98 mJ; magnetic torque performs corresponding positive work.

Check: Integrating signed torque from $\frac{2\pi}{3}$ to $\frac\pi6$ gives the same work. Full antiparallel-to-parallel difference is $2\mu B=60.0\,\mathrm{mJ}$ , which is larger. This does not constitute a complete motor balance of supply, self-field energy and friction.

Why alignment alone does not give continuous motor rotation

A constant-current loop in an unchanged field has an aligning, angle-dependent torque. A full revolution returns to the same orientation potential and supplies no unlimited mechanical energy. A motor maintains its desired rotation through suitable commutation or controlled changing fields. Supply, back emf, load and losses then belong in the balance. Torque temporarily reaching zero does not mean an already moving loop must simultaneously be at rest.[1]

A nonuniform field can produce net force and a different torque distribution. Large nonplanar or differently oriented turns require summing their actual contributions. An iron core also changes field and material response; multiplying air-coil area by an arbitrary enhancement factor does not replace field modelling. The dipole model in a uniform external field is a clear starting point, not a universal description of every coil bearing or electric motor.[1]

Five exercises: angle, axis and multiple moments

Calculate · Foundation

30 turns of area 20.0 cm² each, $I=0.200\,\mathrm A$ , $B=0.500\,\mathrm T$ , angle to normal 30°. Find torque.

Show hint

Convert an area rather than a length.

Learning goal: Find area and torque magnitude in SI.

Show solution and calculation

Approach: $\tau=N|I|AB\sin\theta$ .
Area is 0.00200 m².
The magnitude relation gives 3.00 mN·m.

Check: at ninety degrees it would be 6.00 mN·m.

\tau=(30)(0.200)(0.00200)(0.500)\sin30^\circ\,\mathrm{N\,m}=0.00300\,\mathrm{N\,m}

[1]

Rearrange · Application

At least 0.0120 N·m is required at $I=0.400\,\mathrm A$ , $A=0.00300\,\mathrm{m^2}$ , $B=0.500\,\mathrm T$ and $\theta=60^\circ$ . How many identical turns suffice in the model?

Show hint

Calculate the real target first, then round up.

Learning goal: Design turn count for a minimum magnitude.

Show solution and calculation

Approach: $N_{\min}=\left\lceil\frac{\tau_{\min}}{|I|AB\sin\theta}\right\rceil$ .
Rearrangement gives about 23.094 turns, so 24 are required.

Check: 23 give only 11.951 mN·m, whereas 24 give 12.471 mN·m.

\begin{aligned}N_{\min}&=\left\lceil\frac{0.0120}{(0.400)(0.00300)(0.500)\sin60^\circ}\right\rceil\\&=24\end{aligned}

[1]

Interpret · Application

A fixed dipole in a uniform field is once parallel and once antiparallel. Both torque magnitudes are zero. Are both orientations stable? Explain using potential and a small perturbation.

Show hint

Zero derivative can mark a minimum or maximum.

Learning goal: Distinguish stable and unstable zero-torque states.

Show solution and calculation

Approach: $U(\theta)=-\mu B\cos\theta$ .
Parallel is the minimum and antiparallel the maximum of orientation potential.
Parallel alignment is stable and a small displacement is restored.
Antiparallel alignment is unstable and a small perturbation initially grows.

Check: this does not imply immediate settling without damping.

\begin{aligned}U(0)&=-\mu B\\U(\pi)&=+\mu B\end{aligned}

[1]

Error analysis · Application

A field makes twenty degrees with the loop plane. A solution uses $\sin20^\circ$ in torque magnitude. Correct the angle and find the ratio of correct to wrong magnitude.

Show hint

The normal is perpendicular to the plane.

Learning goal: Convert plane angle to normal angle.

Show solution and calculation

Approach: $\theta=90^\circ-\beta$ .
The normal angle is seventy degrees, so use $\cos20^\circ$ rather than $\sin20^\circ$ .
Correct magnitude is about 2.747 times larger.

Check: a field parallel to the plane must give maximum torque, not zero.

\frac{\tau_{\mathrm{right}}}{\tau_{\mathrm{wrong}}}=\frac{\sin70^\circ}{\sin20^\circ}\approx2.74748

[1]

Transfer · Challenge

Two loops fixed to the same rigid carrier have $\mu_1=0.0200\,\mathrm{A\,m^2}$ along +x and $\mu_2=0.0300\,\mathrm{A\,m^2}$ along +y. Uniform field 0.400 T along +z. Find total torque vector, magnitude and component along an allowed +x rotation axis.

Show hint

The cross product distributes over the sum; the bearing uses only one component.

Learning goal: Add multiple loop moments and project onto an axis.

Show solution and calculation

Approach: $\vec\tau=(\vec\mu_1+\vec\mu_2)\times\vec B$ .
The x-oriented loop gives −y torque and the y-oriented loop +x torque.
Total vector is (0.0120,−0.00800,0) N·m, magnitude 14.422 mN·m; +x axial torque is +12.0 mN·m.

Check: total magnitude differs from axial torque.

\begin{aligned}\vec\tau&=(\vec\mu_1+\vec\mu_2)\times\vec B\\&=(0.0120\hat{\vec x}-0.00800\hat{\vec y})\,\mathrm{N\,m}\\|\vec\tau|&=\sqrt{0.0120^2+0.00800^2}\,\mathrm{N\,m}\end{aligned}

[1]

What to take away

Single-turn area and current circulation determine the dipole vector. Angle to the normal determines torque magnitude; the cross product gives direction. Zero net force does not exclude torque. A bearing uses the permitted axial component, while work requires an actual rotation.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Explained force couple, normal angle and axis projection; wrote two original examples including work balance and five varied DE/EN exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Torque on a current loop. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/drehmoment-stromschleife

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Frequently asked questions about Torque on a current loop

What does Torque on a current loop describe physically?+

A current-carrying loop has magnetic dipole moment $\vec\mu=NIA\hat{\vec n}$ . Its direction follows the right-hand rule for conventional current. In uniform external field, $\vec\tau=\vec\mu\times\vec B$ produces aligning torque. Opposite conductor forces can cancel as net force while their couple rotates the loop. The magnitude formula uses angle between normal and field, not plane and field. A fixed bearing additionally requires projection onto its allowed axis. Area refers to one turn; for nonidentical turns, add individual moment vectors instead of blindly multiplying by one common area. Magnitude alone cannot specify the direction of rotation.

How can I check a calculation involving Torque on a current loop?+

First define the positive normal from current circulation and determine its angle to the field. Convert area to square metres and use one turn’s area. The existing example with one hundred turns, 0.20 amperes, 0.010 square metres and 0.50 tesla gives $0.10\,\mathrm{N\,m}$ at ninety degrees. This is also the maximum magnitude for those inputs; at sixty degrees the result must be smaller. Check direction through the cross product and any necessary axis projection. A check using two opposite conductor forces uses the full separation of their lines of action, not twice that separation. Zero net force remains compatible with nonzero torque.

What assumptions and limits apply to Torque on a current loop?+

The short magnitude formula assumes identical planar turns in uniform external field. The current-oriented normal fixes the dipole vector and angle. With differing areas or orientations, add actual moment vectors. A spatially varying field can produce net force and a different torque distribution; one common field number is then insufficient. The formula also does not automatically give torque along an arbitrary bearing. Project onto the permitted rotation axis for that purpose. Iron cores, field back reaction and changing currents may require an extended model. The simple loop model therefore describes a defined interaction rather than complete operation of an actual motor.

How do I check units and scale for Torque on a current loop?+

Dipole moment has unit $\mathrm{A\,m^2}$ . Multiplying by tesla gives $\mathrm{N\,m}$ , the conventional torque unit. Although this dimension also matches energy, torque is not itself performed work. Work additionally requires rotation and the integral of signed axial torque over angle. Twenty square centimetres equals $0.00200\,\mathrm{m^2}$ ; a linear centimetre conversion would be wrong. As a scale check, torque magnitude in the uniform model cannot exceed $N|I|AB$ . It vanishes when normal is parallel or antiparallel to the field. Unit checking alone detects neither using plane angle instead of normal angle nor selecting an incorrect bearing axis.

When should I combine Torque on a current loop with another relation?+

For rotational motion, supplement electromagnetic torque with mechanical torque balance and an appropriate inertia description. Do not use total vector magnitude as scalar driving torque without checking the axis. At fixed dipole magnitude in a stationary field, $U=-\vec\mu\cdot\vec B$ explains stable parallel and unstable antiparallel alignment. Finite work follows from this potential change or the torque integral. Continuous motor rotation instead requires suitable current reversal or controlled fields and an electrical energy balance. Back emf, supply and losses then belong in the model. Constant aligning-torque modelling supplies no unlimited energy and does not predict immediate quiet settling without damping.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §11.5: Force and Torque on a Current Loop

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\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}

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\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

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Variables & units

Symbol

Meaning

Unit

τ

Torque magnitude

N·m

N

Number of turns

dimensionless

I

Nonnegative current magnitude; normal follows actual conventional circulation

A

Area of one turn

m²

B

Magnitude of uniform external magnetic flux density

θ

Angle between actual dipole vector or current-oriented normal and external field

rad or °

Derivation & background

Exam blueprint

Validity range

Derivation steps

Opposite conductor forces form a couple; total separation of their lines of action gives area dependence.

1Specify identical planar turns and uniform external field.
2The right-hand rule for conventional current defines positive area normal.
3For the chosen rectangle, two opposite sides are transverse to field; their forces are equal and opposite.
4Each force magnitude is $|I|bB$ ; total lever-arm separation is $a\sin\theta$ .
5The couple gives $\tau=|I|abB\sin\theta$ ; with $A=ab$ , N turns add their contributions.
6Define $\vec\mu=NIA\hat{\vec n}$ ; generally $\vec\tau=\vec\mu\times\vec B$ .
7Check normal angle, zero net force and any required projection onto an allowed rotation axis.

How do you calculate with Torque on a current loop?

Here is how to work through a typical Torque on a current loop (τ = N|I|AB sinθ) task step by step:

Task

100 identical turns of area 0.010 m² each carry 0.20 A. Uniform external field is 0.50 T; the current-oriented normal is perpendicular to it. Find torque magnitude.

Solution path

Approach: $\tau=N|I|AB\sin\theta$ ; at $\theta=90^\circ$ , $\tau=0.10\,\mathrm{N\,m}$ .

Which angle and single-turn area determine current-loop torque?

\begin{aligned}\vec\mu&=NIA\hat{\vec n}\\\vec\tau&=\vec\mu\times\vec B\\|\vec\tau|&=N|I|AB\sin\theta\end{aligned}

A force couple gives torque

\begin{aligned}F&=|I|bB\\d_\perp&=a\sin\theta\\\tau&=Fd_\perp=|I|abB\sin\theta\\A&=ab\end{aligned}

Rectangle with sides a and b; the b sides remain transverse to field. $d_\perp$ is the full perpendicular distance between their lines of action.[2][1]

Why is torque zero at both parallel and antiparallel alignment?

Torque magnitude and normal angle

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Original model curve for the 25-turn rectangular coil: this is nonnegative magnitude rather than signed axial torque. The two zeros differ in stability.[1]

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\begin{aligned}U(\theta)&=-\mu B\cos\theta\\\tau_\theta&=-\frac{\mathrm dU}{\mathrm d\theta}=-\mu B\sin\theta\end{aligned}

Ampere square metres and shaft torque

\begin{aligned}[\mu]&=\mathrm{A\,m^2}\\{}[\tau]&=\mathrm{N\,m}\\N&=\frac{\tau}{|I|AB\sin\theta}\\\tau_{\mathrm{axis}}&=\vec\tau\cdot\hat{\vec e}_{\mathrm{axis}}\end{aligned}

Torque is expressed in N·m despite sharing energy’s dimension; turn count must be physically attainable as an integer.[1]

Determine the angle using a weight

Given: A supported coil has fixed dipole magnitude $\mu=0.0800\,\mathrm{A\,m^2}$ in uniform $B=0.300\,\mathrm T$ . A weight acts through an ideal lever with constant perpendicular arm $a=0.0400\,\mathrm m$ : $m=0.0300\,\mathrm{kg}$ , $g=9.81\,\mathrm{m\,s^{-2}}$ . Both torques act about the same axis; other torques vanish. Find an orientation in the first quadrant.
Find: Find the angle to the current-oriented normal for torque balance.

Approach: balance magnetic axial torque against the opposing weight torque. The first-quadrant condition selects one of two possible angles.
$\mu B\sin\theta=mga$
Substitution: first calculate the dimensionless sine, then the angle.
$\begin{aligned}\sin\theta&=\frac{(0.0300)(9.81)(0.0400)}{(0.0800)(0.300)}=0.4905\\\theta&=\arcsin(0.4905)\approx29.37^\circ\end{aligned}$

Result: The angle is approximately 29.4°, with torque magnitude 11.772 mN·m.

Example: check area and lever arms

A tilted rectangular coil

Given: 25 identical turns with $a=0.120\,\mathrm m$ and $b=0.0800\,\mathrm m$ ; current 0.300 A, uniform external field 0.400 T, angle to current-oriented normal 60°. The b sides are transverse to field.
Find: Find dipole magnitude and torque; check with the force couple.

Approach: first find single-turn area, then moment and cross-product magnitude.
$\begin{aligned}A&=ab\\\mu&=NIA\\\tau&=\mu B\sin\theta\end{aligned}$
Substitution: each turn has area 0.00960 m².
$\begin{aligned}\mu&=(25)(0.300)(0.00960)\,\mathrm{A\,m^2}=0.0720\,\mathrm{A\,m^2}\\\tau&=(0.0720)(0.400)\sin60^\circ\,\mathrm{N\,m}\\&\approx0.0249415\,\mathrm{N\,m}\end{aligned}$

Result: Dipole magnitude 0.0720 A·m²; torque magnitude about 24.9 mN·m.

Example: work through a finite angle

Alignment without constant torque

Given: Fixed magnetic dipole magnitude 0.120 A·m² in uniform stationary 0.250 T field. θ decreases from 120° to 30°. Current and geometry are controlled constant; consider the effective mechanical dipole potential change.
Find: Calculate potential change and work by magnetic torque.

Approach: for a fixed dipole, work equals negative change of its orientation potential.
$\begin{aligned}\Delta U&=-\mu B(\cos\theta_f-\cos\theta_i)\\W_{\mathrm{mag}}&=-\Delta U\end{aligned}$
Substitution: retain angular dependence throughout rotation.
$\begin{aligned}\Delta U&=-(0.120)(0.250)(\cos30^\circ-\cos120^\circ)\,\mathrm J\\&\approx-0.0409808\,\mathrm J\\W_{\mathrm{mag}}&\approx+0.0409808\,\mathrm J\end{aligned}$

Result: Orientation potential decreases by 40.98 mJ; magnetic torque performs corresponding positive work.

Why alignment alone does not give continuous motor rotation

Five exercises: angle, axis and multiple moments

Calculate · Foundation

30 turns of area 20.0 cm² each, $I=0.200\,\mathrm A$ , $B=0.500\,\mathrm T$ , angle to normal 30°. Find torque.

Show hint

Convert an area rather than a length.

Learning goal: Find area and torque magnitude in SI.

Show solution and calculation

Approach: $\tau=N|I|AB\sin\theta$ .
Area is 0.00200 m².
The magnitude relation gives 3.00 mN·m.

Check: at ninety degrees it would be 6.00 mN·m.

\tau=(30)(0.200)(0.00200)(0.500)\sin30^\circ\,\mathrm{N\,m}=0.00300\,\mathrm{N\,m}

[1]

Rearrange · Application

At least 0.0120 N·m is required at $I=0.400\,\mathrm A$ , $A=0.00300\,\mathrm{m^2}$ , $B=0.500\,\mathrm T$ and $\theta=60^\circ$ . How many identical turns suffice in the model?

Show hint

Calculate the real target first, then round up.

Learning goal: Design turn count for a minimum magnitude.

Show solution and calculation

Approach: $N_{\min}=\left\lceil\frac{\tau_{\min}}{|I|AB\sin\theta}\right\rceil$ .
Rearrangement gives about 23.094 turns, so 24 are required.

Check: 23 give only 11.951 mN·m, whereas 24 give 12.471 mN·m.

\begin{aligned}N_{\min}&=\left\lceil\frac{0.0120}{(0.400)(0.00300)(0.500)\sin60^\circ}\right\rceil\\&=24\end{aligned}

[1]

Interpret · Application

A fixed dipole in a uniform field is once parallel and once antiparallel. Both torque magnitudes are zero. Are both orientations stable? Explain using potential and a small perturbation.

Show hint

Zero derivative can mark a minimum or maximum.

Learning goal: Distinguish stable and unstable zero-torque states.

Show solution and calculation

Approach: $U(\theta)=-\mu B\cos\theta$ .
Parallel is the minimum and antiparallel the maximum of orientation potential.
Parallel alignment is stable and a small displacement is restored.
Antiparallel alignment is unstable and a small perturbation initially grows.

Check: this does not imply immediate settling without damping.

\begin{aligned}U(0)&=-\mu B\\U(\pi)&=+\mu B\end{aligned}

[1]

Error analysis · Application

A field makes twenty degrees with the loop plane. A solution uses $\sin20^\circ$ in torque magnitude. Correct the angle and find the ratio of correct to wrong magnitude.

Show hint

The normal is perpendicular to the plane.

Learning goal: Convert plane angle to normal angle.

Show solution and calculation

Approach: $\theta=90^\circ-\beta$ .
The normal angle is seventy degrees, so use $\cos20^\circ$ rather than $\sin20^\circ$ .
Correct magnitude is about 2.747 times larger.

Check: a field parallel to the plane must give maximum torque, not zero.

\frac{\tau_{\mathrm{right}}}{\tau_{\mathrm{wrong}}}=\frac{\sin70^\circ}{\sin20^\circ}\approx2.74748

[1]

Transfer · Challenge

Two loops fixed to the same rigid carrier have $\mu_1=0.0200\,\mathrm{A\,m^2}$ along +x and $\mu_2=0.0300\,\mathrm{A\,m^2}$ along +y. Uniform field 0.400 T along +z. Find total torque vector, magnitude and component along an allowed +x rotation axis.

Show hint

The cross product distributes over the sum; the bearing uses only one component.

Learning goal: Add multiple loop moments and project onto an axis.

Show solution and calculation

Approach: $\vec\tau=(\vec\mu_1+\vec\mu_2)\times\vec B$ .
The x-oriented loop gives −y torque and the y-oriented loop +x torque.
Total vector is (0.0120,−0.00800,0) N·m, magnitude 14.422 mN·m; +x axial torque is +12.0 mN·m.

Check: total magnitude differs from axial torque.

\begin{aligned}\vec\tau&=(\vec\mu_1+\vec\mu_2)\times\vec B\\&=(0.0120\hat{\vec x}-0.00800\hat{\vec y})\,\mathrm{N\,m}\\|\vec\tau|&=\sqrt{0.0120^2+0.00800^2}\,\mathrm{N\,m}\end{aligned}

[1]

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Explained force couple, normal angle and axis projection; wrote two original examples including work balance and five varied DE/EN exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Torque on a current loop. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/drehmoment-stromschleife

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Create a curated FSRS exam set for τ = N|I|AB sinθ: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Torque on a current-carrying loop

Variables & units

Current loop torque calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetism

Worked example

Applications

How do you calculate with Torque on a current loop?

Which angle and single-turn area determine current-loop torque?

A force couple gives torque

Why is torque zero at both parallel and antiparallel alignment?

Torque magnitude and normal angle

Ampere square metres and shaft torque

Determine the angle using a weight

Example: check area and lever arms

A tilted rectangular coil

Example: work through a finite angle

Alignment without constant torque

Why alignment alone does not give continuous motor rotation

Five exercises: angle, axis and multiple moments

30 turns of area 20.0 cm2 each, I=0.200 AI=0.200\,\mathrm AI=0.200A, B=0.500 TB=0.500\,\mathrm TB=0.500T, angle to normal 30°. Find torque.

At least 0.0120 N·m is required at I=0.400 AI=0.400\,\mathrm AI=0.400A, A=0.00300 m2A=0.00300\,\mathrm{m^2}A=0.00300m2, B=0.500 TB=0.500\,\mathrm TB=0.500T and θ=60∘\theta=60^\circθ=60∘. How many identical turns suffice in the model?

A fixed dipole in a uniform field is once parallel and once antiparallel. Both torque magnitudes are zero. Are both orientations stable? Explain using potential and a small perturbation.

A field makes twenty degrees with the loop plane. A solution uses sin⁡20∘\sin20^\circsin20∘ in torque magnitude. Correct the angle and find the ratio of correct to wrong magnitude.

What to take away

Sources for this explanation

Frequently asked questions about Torque on a current loop

Subject references and further reading

More formulas: Electricity and magnetism

Retain Torque on a current loop for exams

Torque on a current-carrying loop

Variables & units

Current loop torque calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetism

Worked example

Applications

How do you calculate with Torque on a current loop?

Which angle and single-turn area determine current-loop torque?

A force couple gives torque

Why is torque zero at both parallel and antiparallel alignment?

Torque magnitude and normal angle

Ampere square metres and shaft torque

Determine the angle using a weight

Example: check area and lever arms

A tilted rectangular coil

Example: work through a finite angle

Alignment without constant torque

Why alignment alone does not give continuous motor rotation

Five exercises: angle, axis and multiple moments

30 turns of area 20.0 cm2 each, I=0.200 AI=0.200\,\mathrm AI=0.200A, B=0.500 TB=0.500\,\mathrm TB=0.500T, angle to normal 30°. Find torque.

At least 0.0120 N·m is required at I=0.400 AI=0.400\,\mathrm AI=0.400A, A=0.00300 m2A=0.00300\,\mathrm{m^2}A=0.00300m2, B=0.500 TB=0.500\,\mathrm TB=0.500T and θ=60∘\theta=60^\circθ=60∘. How many identical turns suffice in the model?

A fixed dipole in a uniform field is once parallel and once antiparallel. Both torque magnitudes are zero. Are both orientations stable? Explain using potential and a small perturbation.

A field makes twenty degrees with the loop plane. A solution uses sin⁡20∘\sin20^\circsin20∘ in torque magnitude. Correct the angle and find the ratio of correct to wrong magnitude.

What to take away

Sources for this explanation

Frequently asked questions about Torque on a current loop

Subject references and further reading

More formulas: Electricity and magnetism

Retain Torque on a current loop for exams

30 turns of area 20.0 cm² each, $I=0.200\,\mathrm A$ , $B=0.500\,\mathrm T$ , angle to normal 30°. Find torque.

At least 0.0120 N·m is required at $I=0.400\,\mathrm A$ , $A=0.00300\,\mathrm{m^2}$ , $B=0.500\,\mathrm T$ and $\theta=60^\circ$ . How many identical turns suffice in the model?

A field makes twenty degrees with the loop plane. A solution uses $\sin20^\circ$ in torque magnitude. Correct the angle and find the ratio of correct to wrong magnitude.

30 turns of area 20.0 cm² each, $I=0.200\,\mathrm A$ , $B=0.500\,\mathrm T$ , angle to normal 30°. Find torque.

At least 0.0120 N·m is required at $I=0.400\,\mathrm A$ , $A=0.00300\,\mathrm{m^2}$ , $B=0.500\,\mathrm T$ and $\theta=60^\circ$ . How many identical turns suffice in the model?

A field makes twenty degrees with the loop plane. A solution uses $\sin20^\circ$ in torque magnitude. Correct the angle and find the ratio of correct to wrong magnitude.