Q.Obtain an expression for the torque on a rectangular loop placed in a uniform magnetic field. OR Obtain an expression for the magnetic dipole moment of a revolving electron.
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Torque on a Current Loop
The Intuition First
Imagine a compass needle in the Earth's magnetic field. The needle always turns until it points north. Why? Because the needle itself is a tiny magnet, and the field exerts a twist — a torque — that tries to align it.
A current-carrying loop behaves exactly like that tiny magnet. It has a magnetic moment m, which is like its own internal compass arrow. When you place this loop in an external magnetic field B, the field pulls on one side of the loop and pushes on the other, creating a turning effect.
The loop doesn't feel a net force (if the field is uniform), but it does feel a torque. That torque always tries to rotate the loop so that its magnetic moment points along the field — just like a compass needle.
The Key Players
The magnetic moment of a planar current loop is:
m=IAn^
where I is the current, A is the area of the loop, and n^ is a unit vector perpendicular to the plane of the loop (direction given by the right-hand rule: curl your fingers along the current, your thumb points along m).
The external field B is uniform — same magnitude and direction everywhere in the region of the loop.
The Torque: Two Equivalent Forms
The torque on the loop is:
τ=mBsinθ
where θ is the angle between m and B. The torque is maximum when m is perpendicular to B (θ=90∘), and zero when they are parallel or antiparallel (θ=0∘ or 180∘).
The vector form captures both magnitude and direction:
τ=m×B
τ=m×B
The cross product tells you: the torque is perpendicular to both m and B, and its direction is given by the right-hand rule. This torque always rotates m toward B.
Why It Happens (The Physics)
Consider a rectangular loop of sides a and b, carrying current I, placed in a uniform field B. Let the plane of the loop make an angle θ with the field.
The two sides of length a are perpendicular to B. On each of these sides, the magnetic force is F=IaB, but the forces on opposite sides are in opposite directions. These two forces form a couple — equal and opposite, not along the same line — which produces a torque.
The lever arm for each force is (b/2)sinθ, so the net torque is:
τ=2×(IaB)×2bsinθ=I(ab)Bsinθ=IABsinθ
Since m=IA, we get τ=mBsinθ.
For a rectangular loop, the torque comes only from the sides perpendicular to the field. The sides parallel to the field experience forces that are either zero or along the axis — they contribute nothing to the torque.
The Stable Equilibrium
When m is aligned with B (θ=0), the torque is zero. This is a stable equilibrium — if you nudge the loop slightly, the torque brings it back. …
Torque on a current loop: τ=m×B, τ=NIABsinθ (from the forces on the loop's sides).
OR: μl=2meeL=2evr (magnetic moment of a revolving electron). …
The torque on a current loop in a uniform field follows from adding the moments of the forces on its four sides, giving τ=NIABsinθ=∣m×B∣. In the alternative, treating the orbiting electron as a tiny current loop gives its magnetic moment as μl=evr/2=(e/2me)L.
Torque on a rectangular current loop
Consider a rectangular loop PQRS of sides a and b, carrying current I, placed in a uniform field B with its normal making angle θ with B.
The two sides of length b, perpendicular to B's in-plane component, experience forces F=BIb each (from F=Il×B), equal, opposite and non-collinear — forming a couple. The two sides of length a experience equal, opposite, collinear forces (along the loop's axis), giving no net torque.
The perpendicular distance between the two forces F=BIb (moment arm) is asinθ (the loop's plane is tilted at θ to B). So:
τ=F×(arm)=(BIb)(asinθ)=BI(ab)sinθ=BIAsinθ
with A=ab. For N turns:
τ=NIABsinθ
Defining the loop's magnetic moment m=NIA (along the normal, right-hand rule), compactly:
τ=m×B
…
- CBSE 2026Set ANNUAL1 markMCQQ.If vector m be the magnetic moment of a magnetic dipole placed in a magnetic field of induction vector B, the torque experienced by the dipole will be(a) m . B(b) |m| / |B|(c) m x B(d) |m| |B|
›Reveal solutionSolution
The torque on a magnetic dipole is τ=m×B, analogous to torque on an electric dipole p×E.
When a magnetic dipole of moment m is placed in a uniform magnetic field B, it experiences a torque that tends to align it with the field. This torque is given by the vector product:
τ=m×B
…
- CBSE 2026Set SEM31 markMCQQ.The ratio of the radii of two circular loops is 1 : 2. The ratio of their magnetic moments is 1 : 2. The ratio of currents flowing through them is(a) 1 : 1(b) 2 : 1(c) 4 : 1(d) 1 : 4
›Reveal solutionSolution
Magnetic moment M = I·(πr²), so I = M/(πr²) ∝ M/r². Substituting the given ratios gives I₁ : I₂ = 2 : 1. Option (b).
Step 1 — magnetic moment of a current loop (NCERT/CBSE Class 12 Physics, Moving Charges and Magnetism): M = I·A = I·πr².
Step 2 — so current I = M/(πr²), i.e. I ∝ M/r².
…
- CBSE 2025Set ANNUAL1 markMCQQ.Two circular loops having ratio of their radii 1 : 2 possess same magnetic moment. The ratio of their circulating currents will be(a) 4 : 1(b) 1 : 4(c) 2 : 1(d) 1 : 2
›Reveal solutionSolution
Magnetic moment m = IA = Iπr²; equal m with r ratio 1:2 forces the current ratio to be 4:1 (inverse of the area ratio).
Magnetic moment of a current loop is m=Iπr2. Let the radii be r1:r2=1:2 and the moments be equal, m1=m2: …
- CBSE 2024Set A11 markQ.The torque on a rectangular current loop in a uniform magnetic field increases by ———————— the area of the loop. Fill in the blank choosing the appropriate answer from the bracket: (decreasing, interference, helium, greater, diffraction, increasing)
›Reveal solutionSolution
increasing (the torque is directly proportional to the area of the loop). …
- CBSE 2024Set A1 markMCQQ.The value of torque (τ) experienced by current loop of magnetic moment (m) placed in magnetic field (B) is (A) τ = m × B (B) τ = B × m (C) τ = m/B (D) τ = B/m
›Reveal solutionSolution
A current loop behaves like a magnetic dipole; the torque on it is the cross product of its magnetic moment and the field, τ = m × B.
A planar current loop carrying current I and enclosing area A has a magnetic (dipole) moment m=IA, directed along the normal to the loop (right-hand rule).
When this dipole is placed in a uniform magnetic field B, the two sides of the loop carry equal and opposite forces that form a couple. The resulting torque is
τ=m×B,τ=mBsinθ
…
- CBSE 2024Set ANNUAL1 markQ.Write the formula for torque acting on rotating current carrying coil in terms of magnetic dipole moment, in vector form.
›Reveal solutionSolution
Torque on a magnetic dipole (current loop) in a magnetic field.
A current-carrying coil of magnetic dipole moment m placed in a uniform magnetic field B experiences a torque that tends to align m with B:
τ=m×B …
- CBSE 2023Set 55/4/11 markMCQQ.Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false. Assertion (A) : A current carrying square loop made of a wire of length L is placed in a magnetic field. It experiences a torque which is greater than the torque on a circular loop made of the same wire carrying the same current in the same magnetic field. Reason (R) : A square loop occupies more area than a circular loop, both made of wire of the same length.
›Reveal solutionSolution
For a fixed wire length L, a circle encloses the maximum area, so the square loop has less area than the circular loop. Since torque τ=NIABsinθ depends directly on area, the square loop experiences less torque — making Assertion false and Reason false as well. The correct choice is (D).
The Core Idea: Magnetic Torque and Area
When a current-carrying loop sits in a uniform magnetic field, the net force on it is zero, but the field exerts a torque that tries to rotate the loop. The magnitude of this torque is given by:
τ=NIABsinθ
Here N is the number of turns (1 for a single loop), I is the current, A is the area enclosed by the loop, B is the magnetic field strength, and θ is the angle between the loop’s normal and the field. For a fixed I, B, and θ, the torque is directly proportional to the area A.
So the question boils down to: for a given wire length L, which shape — square or circle — gives a larger area?
Step-by-Step Reasoning
- The wire length is fixed. Both loops are made from the same wire of total length L. This length becomes the perimeter of each loop. For the square, each side is L/4, so its area is:
Asquare=(4L)2=16L2
- For the circle, the circumference is L=2πr, so the radius is r=L/(2π). The area is:
Acircle=πr2=π(2πL)2=4πL2
- Compare the two areas. Which is larger? Compare L2/16 and L2/(4π). Since L2 is positive, we compare the denominators: 16 vs 4π≈12.57. A smaller denominator means a larger fraction, so:
4πL2>16L2
Therefore, Acircle>Asquare.
TipThis is a classic isoperimetric result: for a given perimeter, the circle encloses the maximum possible area. Any other shape — square, rectangle, triangle — will have a smaller area. Memorising this saves you from recalculating every time. …
- CBSE 2019Set ANNUAL1 markMCQQ.The torque (τ⃗) experienced by a current - loop of magnetic moment (M⃗) placed in magnetic field B⃗ is -(a) τ⃗ = M⃗ × B⃗(b) τ⃗ = B⃗ × M⃗(c) τ⃗ = M⃗ / B⃗(d) τ⃗ = M⃗ . B⃗
›Reveal solutionSolution
Torque on a current loop: τ⃗ = M⃗ × B⃗.
A current loop of magnetic moment M⃗ = NIA n̂ placed in a uniform magnetic field B⃗ experiences a torque
τ=M×B,
…
- CBSE 2016Set ANNUAL1 markMCQQ.A conducting circular loop of radius r carries a constant current I. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is(a) BIr(b) 2πrIB(c) zero(d) πrIB
›Reveal solutionSolution
The net magnetic force on any closed current loop placed in a uniform magnetic field is always zero.
For a current loop in a UNIFORM magnetic field B, the net force is F = I∮dl×B = I(∮dl)×B. Since the vector sum of all the line elements around a closed loop is zero (∮dl = 0), the net force is always zero — regardless of the loop's shape or its orientation relative to B. (A net torque can still act if …
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