Force and torque on a current loop
Show why uniform fields give zero net force but nonzero torque on current loops, and use tau = mu cross B confidently.
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Guided steps
- A closed path has zero summed directed length, so a uniform field gives zero net force. Turning is a different sum: equal opposite forces on separated lines can have moments in the same direction. The zero force does not decide torque.
- Curl your right-hand fingers with conventional current to choose the normal. Number of turns times current times area gives the magnetic moment along that normal. The plane and normal are perpendicular, so a plane angle cannot be inserted unchanged into the torque sine. Current direction also chooses the normal needed for an energy sign.
- Each of the two perpendicular wire sides feels current times side length times field. The two half lever arms add to the perpendicular separation, a sine theta. This produces current times area times field times sine of the normal-field angle. In the figure the upper force is up and the lower is down on different lines: they turn the normal clockwise towards the rightward field. Use moment cross field for the torque vector.
- Torque is a cross product, while orientational energy is minus a dot product. Both aligned and anti-aligned moments have zero torque. Alignment is an energy minimum and restoring; anti-alignment is an energy maximum and unstable. A loop without damping can pass through alignment rather than stop there.
- The original tilted-coil example gives the normal angle directly. Form the magnetic moment once, then multiply by field and the sine of that normal angle. Check the answer against the maximum torque; the magnitude alone does not specify an axis.
- Now the given angle is to the plane. Choose the current-selected normal first, including the supplied acute-angle information. Keep moment magnitude, torque magnitude and orientational energy separate, and compare each with its own unit and limit.
- Find the moment magnitude, torque magnitude and signed orientational energy. Choose the angle before substituting; explain that choice separately.
- Explain which angle the normal makes and why the energy has its sign. Compare your argument yourself. Checking three values does not mark the angle or energy reasoning; paper work is fine.
- Use fixed axes rather than an angle shortcut now. Infer the normal from the stated current, then take moment cross field and minus moment dot field. Separately add forces around the closed loop. A field component can contribute to one calculation while dropping out of another.
- Find signed y torque, orientational energy and the net-force choice in the two-component field. Use the current-selected normal and right-handed axes.
- Explain which field components enter torque and energy, why the net force is zero, and what reversing current changes at the same physical orientation. Compare the vector and system-boundary arguments yourself; the checker does not mark them.
- Predict the torque when the current-selected normal is parallel to the field. Keep zero torque separate from the general question of stability.
- Predict maximum torque after doubling turns while current, area per turn and field stay fixed. Do not silently hold total wire length or electrical supply voltage fixed instead.
- Calculate torque when the normal is perpendicular to the field. Check the orientation before using the maximum value.
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A closed current loop can turn even when its net force is zero. To explain that motion, keep two sums separate: add forces to decide translation, and add their moments to decide turning. The loop’s current gives its area a direction; that directed area connects the individual forces to a magnetic dipole moment.
Use the current-element force law and the distinction between a force and its moment. Here the loop is planar and rigid, the current is steady, and the external magnetic field is uniform over every turn. The field made by the loop itself is not the field inserted in its own torque law.
Separate zero net force from a turning couple
For a closed loop carrying constant current I in one uniform field,
The directed length elements sum to zero because the path returns to its starting point. Individual wire elements can still feel forces. Equal opposite forces on different lines of action form a couple: their vector sum is zero, but their turning moments can add. Thus zero net force is insufficient information to infer zero torque. This closed-loop force result needs a uniform field; in a varying field the forces need not cancel.
Let the current choose the area normal
Curl your right-hand fingers along conventional current around the loop. Your thumb defines the unit normal n hat. For N identical coplanar turns of area A each, define the magnetic dipole moment
Its SI unit is A m². Here I is a positive current magnitude; its direction has already been encoded in the normal. Reversing current reverses μ. Increasing N multiplies the moment only when the current and area per turn stay fixed.
Define θ as the angle between μ and B, from 0 to π. The plane of the coil is perpendicular to its normal. If a problem gives an acute angle α between the plane and the field, the torque magnitude uses sin θ = cos α. A plane angle alone does not choose the signed normal for an energy or vector-torque calculation: the current direction must do that.
Derive the torque from the moment arms
Take a rectangular turn with one pair of sides of length b perpendicular to B. The other sides have length a, so A = ab. Each b side feels a force of magnitude IbB; these two forces are opposite. Their perpendicular separation is a sin θ, not always a. About the centre, the two lever arms are each a sin θ/2. Their torques have the same turning sense, giving
The other pair’s force lines are collinear and their total torque is zero. Add the identical turns to obtain
The vector law supplies the axis and turning sense. A positive torque magnitude alone does not. For example, μ along + z crossed with B along + x gives torque along + y. The law also holds for other planar loop shapes in a uniform external field; the rectangle makes its mechanical origin visible. These force, moment and energy results are developed in OpenStax’s treatment of current-loop torque.
Use energy to distinguish the two zero-torque orientations
For fixed magnetic-moment magnitude and uniform external field, the orientational potential energy is
with U = 0 at θ = 90°. Indeed, the torque component tending to change θ is -dU/dθ = -μ B sin θ. This sign makes a small displacement from alignment restoring.
- At θ = 0, torque is zero and U = -μ B is a minimum: alignment is stable.
- At θ = π, torque is also zero, but U = +μ B is a maximum: anti-alignment is unstable. Zero torque does not by itself mean stable equilibrium.
- At θ = 90°, torque magnitude is maximum and U = 0, not minimum.
A freely moving loop need not stop at alignment; without damping it can overshoot and oscillate. For a slow controlled reorientation at fixed current and field, external work against the magnetic torque equals Δ U. This is an orientational-energy description, not a claim that magnetic force on an individual charge supplies energy; maintaining a motor’s current also involves the electrical source.
Torque on a tilted coil
Worked example 1
Torque from the area normal
Problem
A rectangular coil has N = 20, I = 0.50 A and area A = 0.010 m² in a uniform field B = 0.20 T. Its normal makes 30° with the field. Find the torque magnitude.
Show full solution
Choose the angle and calculate torque
Method
Use τ = NIAB sin θ with θ = 30°.Reason
The magnetic moment points along the area normal, so the stated angle is already the angle in the torque law.Working
Answer
τ = NIAB sin θ = (20)(0.50)(0.010)(0.20) sin 30°.τ = (0.10)(0.20)(0.5) = 1.0 × 10⁻² N m.State the torque
Working
Answer
The torque magnitude is 1.0 × 10⁻² N m. It acts to align the magnetic moment with the field.
Quick check
The torque is largest at 90°, 2.0 × 10⁻² N m, and zero when the normal lies along the field.
Read an angle given to the plane
Guided practice 1
The plane angle is not the normal angle
Problem
A rigid coil has N = 30, current I = 0.40 A and area A = 0.010 m² per turn in a uniform external field B = 0.20 T. The plane makes 30° with the field; the current-selected normal makes an acute angle with the field. Find the magnetic-moment magnitude, torque magnitude and orientational potential energy relative to U = 0 when the normal is perpendicular to the field. Explain your angle choice.
Choose the directed normal before calculating
Hints
Hint 1
Draw the normal
The normal is perpendicular to the plane. Use the stated acute normal to select the angle and energy sign.Hint 2
Separate moment, torque and energy
First form N times I times area. Then use a cross product for torque and a negative dot product for energy.Show solution step by step
Choose the normal angle and moment
Method
Use θ = 90°-30° = 60° and μ = NIA.Reason
The torque and energy laws use the current-selected normal, not the plane.Working
Answer
μ = (30)(0.40)(0.010) = 0.12 A m².Calculate two different physical quantities
Method
Use the sine for torque and the negative cosine for energy.Reason
The cross product measures the turning tendency; the dot product gives the orientation’s energy.Working
Answer
τ = (0.12)(0.20) sin 60° = 2.1 × 10⁻² N m and U = -(0.12)(0.20) cos 60° = -1.2 × 10⁻² J. The torque is below its 0.024 N m maximum. Compare the angle argument and energy sign yourself; checking the numbers does not mark that explanation.
Quick check
Decide translation, turning and energy separately
Check your understanding 1
A coil with N = 20, I = 0.50 A and area A = 0.012 m² per turn lies in the x-y plane. Its conventional current is anticlockwise when viewed from + z looking towards the coil. The uniform external field is B = (0.30x hat + 0.40z hat) T. Use right-handed axes. Find the y component of torque, the orientational potential energy and the net-force magnitude. Explain which field component causes torque, and how reversing the current at the same orientation changes torque and energy.
Show hint
Show answer
The current gives μ = +0.12z hat A m². Thus τ = (0.12)(0.30)y hat = +0.036y hat N m and U = -(0.12)(0.40) = -0.048 J. The parallel z field gives no torque; the perpendicular x field gives no dot-product contribution. The net force is zero because the loop is closed and the field uniform. Reversing current reverses μ, so τy = -0.036 N m and U = +0.048 J; net force remains zero. Compare these vector and system-boundary arguments yourself; the checker verifies the requested values and force choice only.
Check your understanding 2
A current loop’s normal is parallel to a uniform magnetic field. What is its torque?
Show hint
Identify the angle between the magnetic moment and the field.
Show answer
τ = μ B sin 0° = 0 N m. The normal and magnetic moment are aligned with the field.
Check your understanding 3
If a coil’s number of turns doubles while its current, area per turn and field strength stay fixed, what happens to its maximum torque?
Show hint
Write the torque at the angle for which sin θ = 1.
Show answer
It doubles: τ max = NIAB is proportional to N when I, A and B are fixed.
Check your understanding 4
A coil has N = 50, I = 0.20 A, A = 5.0 × 10⁻³ m² and B = 0.40 T. Its normal is perpendicular to the field. Find its torque magnitude.
Show hint
The stated angle makes the sine factor equal to one.
Show answer
τ = NIAB sin 90° = (50)(0.20)(5.0 × 10⁻³)(0.40) = 2.0 × 10⁻² N m. Perpendicular refers to the normal, so this is maximum torque.
Keep force, torque and stability distinct
Choose the current-selected normal before the angle. Sum forces to decide translation, use μ × B for turning, and use -μ · B to distinguish stable alignment from unstable anti-alignment. A zero result in one calculation cannot stand in for the other two.
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