Read this once slowly. Every single question in Part 3 comes from one of these ideas.
1. What is torque?
Torque is the physics word for turning effect. It is what you apply to a door handle, a spanner, or a steering wheel. Notice something about a steering wheel: your two hands push in opposite directions, so the wheel does not move anywhere — but it turns.
That is the heart of this whole topic. A current-carrying loop placed in a magnetic field behaves exactly like that steering wheel.
The one-line summary of section 4.9:
A current loop in a uniform magnetic field feels no net force — it does not get dragged anywhere — but it does feel a torque, so it spins on the spot until it lines up with the field.
Two equal, opposite pushes on different lines. The wheel goes nowhere — it just turns. That turning effect is torque.
2. Why is there no net force?
Take a rectangular loop of wire with sides a and b, carrying current I, sitting in a magnetic field B.
Two of the arms carry current one way; the opposite two carry it the other way.
From F = BIL sin θ, opposite arms therefore feel forces that are equal in size but opposite in direction.
Add them all up: they cancel. Net force = 0.
But the two forces do not act at the same place. One pushes the left edge in, the other pushes the right edge out. A pair of equal, opposite forces acting along different lines is called a couple, and a couple makes things rotate.
Equal and opposite forces always give zero net force. Whether anything turns depends entirely on whether they act along the same line.
Picture it: two children on a merry-go-round, one pushing on each side in opposite directions. The merry-go-round does not slide across the playground — but it spins. Equal and opposite forces on different lines = rotation.
3. Building the torque formula
Case 1 — the field lies flat in the plane of the loop.
Two arms lie along the field, so they feel nothing (sin 0° = 0). The other two arms are at right angles to the field, so each feels the full force F = IbB. Each acts at a distance a/2 from the centre line, and both twist the loop the same way, so they add:
τ = IbB·(a/2) + IbB·(a/2) = I(ab)B = IAB
where A = ab is the area of the loop. Notice what has vanished — the shape does not matter, only the area.
Case 2 — the loop is tilted.
As the loop tilts, the two forces slide closer to being in a straight line with each other, so the turning effect weakens:
τ = I A B sin θ
The trap that decides the mark. The angle θ is measured between the magnetic field B and the normal to the loop — the imaginary skewer sticking straight out of the loop's face. It is not the angle between the field and the loop's surface.
So when the question says "the plane of the coil is parallel to the field", that means θ = 90° and the torque is maximum.
And when it says "the plane of the coil is perpendicular to the field", that means θ = 0° and the torque is zero.
Drag the slider. The loop is drawn edge-on, so it looks like a bar. The field B points right.
Watch what happens to the gap between the two force arrows.
angle θ60°
sin θ0.87
torque τ = mB sinθ0.87 mB
energy U = −mB cosθ−0.50 mB
Notice: at θ = 0° the two arrows line up and the torque dies to zero. At θ = 90° the gap is widest and the torque peaks — and that is the position where the plane of the loop contains the field.
4. The magnetic moment — one number for the whole loop
Instead of writing I and A separately every time, we bundle them into one quantity called the magnetic moment:
m = I A (one turn) m = N I A (N turns)
Think of it as a measure of "how strong a little magnet this loop is." Big current in a big loop = a strong little magnet. Its unit is A m² (ampere metre squared) — literally "current × area", exactly as the definition says. Its direction is the skewer direction, found with the right-hand rule for a loop (curl your fingers along the current, your thumb points along m).
With this, both cases collapse into one beautiful line:
τ = m × B so τ = mB sin θ
Curl your right fingers along the current and your thumb points along m. Everything about the loop is now bundled into that one arrow.
Same shape as Chapter 1. An electric dipole in an electric field obeys τ = pe × E. Swap the electric dipole for a current loop and you get the magnetic version. Physics keeps reusing the same equation.
5. Where does the loop come to rest?
The torque is zero in two positions — when m is parallel to B, and when it is antiparallel. But those two resting places are not equally safe.
Position
θ
Torque
Energy U
What happens if you nudge it
Stable — m along B
0°
0
−mB (minimum)
Torque pushes it back. Like a ball resting in a bowl.
Sideways
90°
mB (maximum)
0
Swings hard towards alignment.
Unstable — m opposite B
180°
0
+mB (maximum)
It flips right over. Like a ball balanced on top of an upturned bowl.
Both start with zero torque. Only one of them stays put. A ball resting in a bowl versus a ball balanced on top of one.
This is also the answer to a question from page one of the chapter: why does a compass needle point north? Because the needle is a magnetic dipole, and torque swings it until it lines up with the Earth's field.
6. Energy and work
Since it takes effort to twist the loop away from its comfortable position, there is a potential energy stored:
U = − m · B = − mB cos θ
And the work needed to turn it from angle θ₁ to angle θ₂ is just the change in that energy:
W = mB (cos θ₁ − cos θ₂)
The loop rolls downhill to θ = 0° and settles there. Lifting it all the way to the top costs W = 2mB — the full height of the hill.
Three results worth memorising outright:
0° → 90° : W = mB
0° → 180° (stable all the way to unstable) : W = 2mB
90° → 180° : W = mB
7. The loop seen from far away — a magnetic dipole
Stand a long way along the axis of the loop (x ≫ R). The loop's radius no longer matters compared with your distance, and the axial field from section 4.5 simplifies to:
B = (μ₀/4π) · (2m / x³)
Two things to notice. First, the field now dies away as 1/x³, not 1/x². Second, I and R have merged into the single quantity m — from far away you cannot tell whether it is a big loop with a small current or a small loop with a big current. Only the product matters.
Far from the loop the field collapses far faster than a wire's 1/r. And from out here you can no longer tell a big loop with a small current from a small loop with a big current — only the product m = IA survives.
Compare the three distance laws in this chapter so they never get mixed up:
Source
Field falls off as
Long straight wire
1 / r
Point charge (electric)
1 / r²
Current loop / dipole, far away
1 / x³
8. Three NCERT reasoning results that come up as MCQs
A uniform field cannot spin a flat loop about its own vertical axis. Because τ = I A × B, and the area vector of a horizontal loop points vertically, the torque must lie in the loop's plane. It can tip the loop over, never spin it like a top.
Stable equilibrium gives maximum flux. When A lines up with B, the loop's own field points the same way as the external field, so the total flux through it is the greatest possible.
A flexible irregular loop springs into a circle. It wants the largest possible area (more area → more moment → more flux), and for a fixed perimeter a circle encloses the greatest area of any shape.
Because τ = I A × B and a cross product is perpendicular to both, the torque on a flat horizontal loop must lie in the horizontal plane. It can tip the loop over; it can never spin it like a top.
The outward forces on every element push the loop open. The wire length can't change, so the only way to gain area is to become a circle, with its plane square-on to the field.
9. The bonus result NEET has started using
An electron going round an orbit is a tiny current loop. If it goes round with speed v at radius r, it completes v/2πr loops each second, so the current is I = ev/2πr. Therefore:
m = I A = (ev/2πr) · πr² = evr / 2
Nothing new is needed here — just the realisation that a charge going round and round is a current. Count how many times it passes a point each second and you have the current.
This one line is what the hardest recent NEET question from this chapter was built on. Worth knowing.
Part 2 — Formula sheet
Everything in section 4.9 on one page. If Aamirah can reproduce this table from memory, she can attempt every question in Part 3.
Quantity
Formula
What to watch out for
Magnetic moment (1 turn)
m = I A
Unit A m². Direction = normal to the loop, by the right-hand rule.
Magnetic moment (N turns)
m = N I A
Forgetting N is the commonest slip.
Area of a circular coil
A = πr²
Convert cm to m before squaring.
Torque (vector form)
τ = m × B
Torque is perpendicular to both m and B.
Torque (magnitude)
τ = m B sin θ = N I A B sin θ
θ is measured from the normal, not from the plane.
Maximum torque
τmax = m B
θ = 90°, i.e. the plane of the coil contains B.
Zero torque
τ = 0
θ = 0° or 180°, i.e. the plane is perpendicular to B.
Net force on a loop
Fnet = 0
True only in a uniform field.
Potential energy
U = − m·B = − mB cos θ
Minimum −mB at θ = 0; maximum +mB at θ = 180°.
Work done rotating
W = mB (cos θ₁ − cos θ₂)
0→90°: W = mB. 0→180°: W = 2mB.
Angular speed after turning
½ Iω² = W ⇒ ω = √(2W/I)
Here I is the moment of inertia, not current.
Far axial field of a loop
B = (μ₀/4π)(2m/x³)
Valid only for x ≫ R. Falls off as 1/x³.
Field at the centre of a coil
B = μ₀NI / 2R
From 4.5 — often combined with m in the same question.
Moment of an orbiting charge
m = e v r / 2 = eL / 2me
An orbiting electron is a current loop.
Electric analogue
τ = pe × E
Substitute μ₀ → 1/ε₀ and m → pe.
Keep these three apart — all contain N, I and R.
Field at the centre of a coil: B = μ₀NI/2R → answer in tesla
Magnetic moment of that coil: m = NIπR² → answer in A m²
Torque on that coil in a field: τ = mB sin θ → answer in N m
Check the unit in the options first. It tells you instantly which formula the question wants.
Shape-changing shortcuts (a favourite NEET trick — same wire, new shape):
Same wire, 1 turn of radius R → n turns: radius becomes R/n, so m → m/n (but the centre field becomes n²B).
Fixed wire length L: circle gives A = L²/4π, square gives A = L²/16 → moments in the ratio 4 : π.
For a fixed perimeter, the circle always gives the largest area, hence the largest moment and torque.
Part 3 — 50 questions with step-by-step solutions
Attempt each one on paper first, then open the solution. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1Moment · basic
A closely wound circular coil of 100 turns and radius 10 cm carries a current of 3.2 A. Its magnetic moment is:
(a) 5 A m²
(b) 10 A m²
(c) 20 A m²
(d) 32 A m²
Show step-by-step solution
GivenN = 100, r = 10 cm = 0.1 m, I = 3.2 A
AskedMagnetic moment m
ConceptA coil of N turns has N times the moment of a single turn. Area of a circle is πr².
Formulam = N I A = N I πr²
SolutionConvert the radius first: r = 10 cm = 0.1 m. A = πr² = 3.14 × (0.1)² = 3.14 × 0.01 = 3.14 × 10⁻² m² m = 100 × 3.2 × 3.14 × 10⁻² m = 320 × 3.14 × 10⁻² = 10.05 A m²
Answer: 10 A m²
Q2Moment · basic
A single square loop of side 10 cm carries a current of 2 A. Its magnetic moment is:
(a) 0.002 A m²
(b) 0.02 A m²
(c) 0.2 A m²
(d) 2 A m²
Show step-by-step solution
GivenSide a = 10 cm = 0.1 m, I = 2 A, N = 1
AskedMagnetic moment m
ConceptFor any flat loop, m = IA. The shape only matters through its area.
Formulam = I A = I a²
SolutionA = a² = (0.1)² = 0.01 m² m = 2 × 0.01 = 0.02 A m²
Answer: 0.02 A m²
Q3Moment · basic
A rectangular coil of 50 turns and dimensions 20 cm × 10 cm carries a current of 1 A. Its magnetic moment is:
(a) 0.1 A m²
(b) 0.5 A m²
(c) 1 A m²
(d) 10 A m²
Show step-by-step solution
GivenN = 50, length = 0.20 m, breadth = 0.10 m, I = 1 A
AskedMagnetic moment m
ConceptSame formula as a circular coil — only the area expression changes.
Formulam = N I A = N I (l × b)
SolutionA = 0.20 × 0.10 = 0.02 m² m = 50 × 1 × 0.02 = 1 A m²
Answer: 1 A m²
Q4Moment · ratio
Two small circular coils carry the same current. Their radii are in the ratio 1 : 2. The ratio of their magnetic moments is:
(a) 1 : 2
(b) 1 : 4
(c) 2 : 1
(d) 4 : 1
Show step-by-step solution
GivenI is the same for both, r₁ : r₂ = 1 : 2
AskedRatio m₁ : m₂
ConceptWith current fixed, m depends only on the area, and area goes as the square of the radius.
Formulam = I πr² ⇒ m ∝ r²
Solutionm₁ / m₂ = r₁² / r₂² = (1)² / (2)² = 1 / 4
Answer: 1 : 4
Q5Moment · ratio
A circular coil has magnetic moment m. If its radius is doubled and the current through it is halved, the new magnetic moment is:
(a) m / 2
(b) m
(c) 2m
(d) 4m
Show step-by-step solution
Givenr → 2r, I → I/2
AskedNew magnetic moment m′
ConceptChange each factor separately, then multiply the effects together.
Formulam = I πr²
SolutionRadius doubled ⇒ area becomes 4 times. Current halved ⇒ multiply by ½. m′ = ½ × 4 × m = 2m
Answer: 2m
Q6Moment · shape
A wire of length L carrying current I is bent first into a circle and then into a square. The ratio of the magnetic moment of the circle to that of the square is:
(a) π : 4
(b) 4 : π
(c) 1 : 1
(d) 2 : π
Show step-by-step solution
GivenSame wire length L, same current I
AskedRatio m_circle : m_square
ConceptThe perimeter is fixed, so work out each area from L, then compare. For a fixed perimeter a circle always encloses more area.
Formulam = IA; circle: 2πr = L; square: 4a = L
SolutionCircle: r = L/2π, so A = πr² = π L²/4π² = L²/4π Square: a = L/4, so A = a² = L²/16 Ratio = (L²/4π) ÷ (L²/16) = 16/4π = 4/π
Answer: 4 : π
Q7Moment · shape
A wire of total length 12a carries a current I. It is bent to form (i) an equilateral triangle of side a and (ii) a square of side a, using the whole wire each time. The ratio of the magnetic moments is:
(a) 1 : √3
(b) √3 : 1
(c) 1 : 3
(d) 3 : 1
Show step-by-step solution
GivenWire length = 12a, side of each shape = a, current I
AskedRatio m_triangle : m_square
ConceptThe number of turns changes with the shape, because the same wire wraps round more times for a smaller perimeter. This is the whole trick.
Formulam = N I A
SolutionTriangle: perimeter = 3a, so N = 12a/3a = 4 turns. Area = (√3/4)a². m₁ = 4 × I × (√3/4)a² = √3 I a² Square: perimeter = 4a, so N = 12a/4a = 3 turns. Area = a². m₂ = 3 × I × a² = 3 I a² Ratio = √3 : 3 = 1 : √3
Answer: 1 : √3
Q8Moment · shape
A wire carrying current I is bent into a single circular loop of radius R with magnetic moment m. The same wire is then re-bent into n identical turns. The new magnetic moment is:
(a) m / n
(b) m / n²
(c) n m
(d) n² m
Show step-by-step solution
GivenSame wire, 1 turn of radius R → n turns, current I unchanged
AskedNew magnetic moment m′
ConceptThe wire length is fixed, so packing n turns shrinks the radius to R/n.
Formulam = N I πr²
SolutionWire length 2πR must equal n × 2πr, so r = R/n. New area = π(R/n)² = πR²/n² m′ = n × I × πR²/n² = (I πR²)/n = m/n
Answer: m / n
Q9Moment · basic
A coil of 200 turns and area 10⁻³ m² carries a current of 0.5 A. Its magnetic moment is:
(a) 0.01 A m²
(b) 0.1 A m²
(c) 1 A m²
(d) 10 A m²
Show step-by-step solution
GivenN = 200, A = 10⁻³ m², I = 0.5 A
AskedMagnetic moment m
ConceptDirect substitution — the area is already given, so no conversion needed.
Formulam = N I A
Solutionm = 200 × 0.5 × 10⁻³ m = 100 × 10⁻³ = 0.1 A m²
Answer: 0.1 A m²
Q10Moment · vector
Two identical circular coils, each of magnetic moment M, are placed concentrically with their planes perpendicular to each other. The resultant magnetic moment is:
(a) 0
(b) M
(c) √2 M
(d) 2M
Show step-by-step solution
GivenTwo coils, each moment M, planes mutually perpendicular
AskedResultant magnetic moment
ConceptMagnetic moment is a vector along the normal. Perpendicular planes mean perpendicular normals, so add them like perpendicular vectors.
Formulam_net = √(m₁² + m₂²)
SolutionThe two moment vectors are at 90° to each other. m_net = √(M² + M²) = √(2M²) m_net = √2 M
Answer: √2 M
Q11Torque · basic
A coil of 50 turns and area 2 × 10⁻² m² carries a current of 5 A. It is placed in a uniform field of 0.5 T with the plane of the coil parallel to the field. The torque on the coil is:
(a) 0
(b) 1.25 N m
(c) 2.5 N m
(d) 5 N m
Show step-by-step solution
GivenN = 50, A = 2 × 10⁻² m², I = 5 A, B = 0.5 T, plane parallel to B
AskedTorque τ
ConceptPlane parallel to B means the normal is perpendicular to B, so θ = 90° and sin θ = 1 — this is the maximum-torque case.
Formulaτ = N I A B sin θ
SolutionPlane ∥ B ⇒ θ (between normal and B) = 90° ⇒ sin θ = 1 τ = 50 × 5 × 2 × 10⁻² × 0.5 × 1 τ = 50 × 5 × 0.02 × 0.5 = 2.5 N m
Answer: 2.5 N m
Q12Torque · basic
The same coil (50 turns, area 2 × 10⁻² m², current 5 A, field 0.5 T) is now placed with its plane perpendicular to the field. The torque is:
(a) 0
(b) 1.25 N m
(c) 2.5 N m
(d) 5 N m
Show step-by-step solution
GivenSame data, but plane perpendicular to B
AskedTorque τ
ConceptPlane perpendicular to B means the normal is along B, so θ = 0°. This is the trap — 'perpendicular' in the question gives θ = 0, not 90°.
Formulaτ = N I A B sin θ
SolutionPlane ⊥ B ⇒ normal is parallel to B ⇒ θ = 0° sin 0° = 0 τ = 0
Answer: 0
Q13Torque · basic
A current loop of magnetic moment 10 A m² is placed in a uniform magnetic field of 2 T such that the angle between the moment and the field is 30°. The torque on the loop is:
(a) 5 N m
(b) 10 N m
(c) 17.3 N m
(d) 20 N m
Show step-by-step solution
Givenm = 10 A m², B = 2 T, θ = 30°
AskedTorque τ
ConceptThe angle is already given between m and B, so substitute directly.
Formulaτ = m B sin θ
Solutionτ = 10 × 2 × sin 30° sin 30° = 0.5 τ = 10 × 2 × 0.5 = 10 N m
Answer: 10 N m
Q14Torque · basic
A magnetic dipole of moment 0.5 A m² is placed at right angles to a uniform field of 0.4 T. The torque acting on it is:
(a) 0
(b) 0.1 N m
(c) 0.2 N m
(d) 0.9 N m
Show step-by-step solution
Givenm = 0.5 A m², B = 0.4 T, θ = 90°
AskedTorque τ
ConceptAt right angles the sine factor is 1, giving the maximum possible torque.
Formulaτ = m B sin θ
Solutionsin 90° = 1 τ = 0.5 × 0.4 × 1 = 0.2 N m
Answer: 0.2 N m
Q15Torque · concept
The torque on a current-carrying loop in a uniform magnetic field is maximum when the plane of the loop is:
(a) perpendicular to the field
(b) parallel to the field
(c) at 45° to the field
(d) torque is the same in all positions
Show step-by-step solution
Givenτ = mB sin θ with θ measured from the normal
AskedOrientation for maximum torque
ConceptMaximum torque needs sin θ = 1, i.e. θ = 90° between the normal and the field. If the normal is at 90° to B, then the plane itself contains B.
Formulaτ = m B sin θ
Solutionτ is maximum when sin θ = 1 ⇒ θ = 90° θ = 90° means the normal is perpendicular to B. If the normal is perpendicular to B, the plane of the loop is parallel to B.
Answer: parallel to the field
Q16Torque · numeric
A coil of area 100 cm² with 20 turns carries a current of 3 A. It is placed in a field of 0.5 T with its normal making 60° with the field. The torque is approximately:
(a) 0.13 N m
(b) 0.26 N m
(c) 0.30 N m
(d) 0.52 N m
Show step-by-step solution
GivenA = 100 cm², N = 20, I = 3 A, B = 0.5 T, θ = 60° (from the normal)
AskedTorque τ
ConceptThe angle is already measured from the normal, so no conversion. Watch the cm² → m² conversion.
Formulaτ = N I A B sin θ
SolutionA = 100 cm² = 100 × 10⁻⁴ = 10⁻² m² m = N I A = 20 × 3 × 10⁻² = 0.6 A m² τ = 0.6 × 0.5 × sin 60° = 0.6 × 0.5 × 0.866 τ = 0.26 N m
Answer: 0.26 N m
Q17Torque · numeric
A circular coil of 10 turns and radius 10 cm carries a current of 2 A. It is placed in a uniform field of 0.2 T with its plane along the field. The torque on the coil is:
(a) 0.063 N m
(b) 0.126 N m
(c) 0.25 N m
(d) 0.63 N m
Show step-by-step solution
GivenN = 10, r = 0.1 m, I = 2 A, B = 0.2 T, plane along B
AskedTorque τ
ConceptPlane along B ⇒ θ = 90° ⇒ sin θ = 1. Find m first, then multiply by B.
Formulaτ = m B, where m = N I πr²
SolutionA = πr² = 3.14 × 0.01 = 3.14 × 10⁻² m² m = 10 × 2 × 3.14 × 10⁻² = 0.628 A m² τ = 0.628 × 0.2 = 0.126 N m
Answer: 0.126 N m
Q18Torque · angle
A current loop of moment m is placed in a field B such that the plane of the loop makes an angle of 30° with the field. The torque on the loop is:
(a) mB / 2
(b) mB √3 / 2
(c) mB
(d) 0
Show step-by-step solution
GivenAngle between the PLANE and B is 30°
AskedTorque τ
ConceptConvert first. If the plane makes 30° with B, the normal makes 90° − 30° = 60° with B. This conversion is the whole question.
Formulaτ = m B sin θ, θ measured from the normal
SolutionPlane makes 30° with B ⇒ normal makes 60° with B τ = mB sin 60° = mB × (√3/2)
Answer: mB √3 / 2
Q19Torque · reverse
A loop of magnetic moment 4 A m² in a field of 0.5 T experiences a torque of 1 N m. The angle between the moment and the field is:
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Show step-by-step solution
Givenm = 4 A m², B = 0.5 T, τ = 1 N m
AskedAngle θ
ConceptRearrange the torque formula to make sin θ the subject.
A given length of wire carrying a fixed current is to be bent into a plane loop and placed in a uniform magnetic field. The torque will be maximum if the loop is:
(a) a square
(b) an equilateral triangle
(c) a circle
(d) a rectangle of any size
Show step-by-step solution
GivenFixed perimeter, fixed current, fixed field
AskedShape giving maximum torque
Conceptτ = IAB sin θ, so with everything else fixed the biggest torque comes from the biggest area. For a fixed perimeter, a circle wins.
Formulaτ = I A B sin θ
SolutionPerimeter is fixed, so we need the shape with the greatest area. For a given perimeter, a circle encloses more area than any other shape. Hence a circular loop gives the maximum torque.
Answer: a circle
Q21Torque · concept
A rectangular current-carrying loop is placed in a uniform magnetic field. The net force on the loop is:
(a) zero
(b) IAB
(c) IAB sin θ
(d) 2IAB
Show step-by-step solution
GivenUniform magnetic field, closed current loop
AskedNet force on the loop
ConceptOpposite arms carry current in opposite directions, so their forces are equal and opposite and cancel. Only a torque survives.
FormulaF = BIL sin θ on each arm; vector sum over the closed loop
SolutionForces on opposite arms are equal in magnitude but opposite in direction. They cancel in pairs when added as vectors. Net force = 0, although the net torque is not zero.
Answer: zero
Q22Torque · concept
A current-carrying circular loop lies on a smooth horizontal table. A uniform magnetic field is applied. The loop can be made to:
(a) spin about its own vertical axis
(b) tip over about a horizontal axis
(c) move sideways across the table
(d) do none of these
Show step-by-step solution
GivenHorizontal loop, uniform field
AskedPossible motion
Conceptτ = I A × B. The area vector of a horizontal loop is vertical, so the torque must lie in the horizontal plane — it can tip the loop but never spin it about the vertical.
Formulaτ = I A × B
SolutionA is vertical for a horizontal loop. A cross product is perpendicular to both vectors, so τ lies in the plane of the loop (horizontal). A horizontal torque tips the loop over; it cannot spin it about the vertical axis. Net force is zero, so it cannot slide either.
Answer: tip over about a horizontal axis
Q23Torque · concept
Two loops, one circular and one square, have the same area and carry the same current. They are placed in the same uniform field at the same orientation. The torques on them are:
(a) equal
(b) greater for the circle
(c) greater for the square
(d) in the ratio 4 : π
Show step-by-step solution
GivenEqual areas, equal currents, same field and orientation
AskedComparison of torques
ConceptTorque depends on the loop only through its area — the shape itself never appears in the formula.
Formulaτ = I A B sin θ
SolutionBoth have the same I, the same A, the same B and the same θ. Therefore both experience exactly the same torque. (Shape matters only when the PERIMETER is fixed instead of the area.)
Answer: equal
Q24Energy · concept
The potential energy of a magnetic dipole of moment m placed in a uniform field B at an angle θ is:
(a) mB sin θ
(b) − mB cos θ
(c) mB cos θ
(d) − mB sin θ
Show step-by-step solution
GivenDipole moment m, field B, angle θ
AskedExpression for potential energy
ConceptThe energy is the negative dot product, exactly like an electric dipole in an electric field.
FormulaU = − m · B
SolutionU = − m·B = − mB cos θ Check: at θ = 0 this gives U = −mB, the lowest energy — which matches the stable position.
Answer: − mB cos θ
Q25Energy · numeric
A magnetic dipole of moment 2 A m² is placed in a uniform field of 0.5 T. The work done in rotating it from the stable position through 180° is:
(a) 0.5 J
(b) 1 J
(c) 2 J
(d) 4 J
Show step-by-step solution
Givenm = 2 A m², B = 0.5 T, rotation 0° → 180°
AskedWork done W
ConceptStable position means θ₁ = 0°. Rotating fully over to 180° is the maximum possible work.
The potential energy of a magnetic dipole is maximum when the dipole moment is:
(a) parallel to B
(b) antiparallel to B
(c) perpendicular to B
(d) at 45° to B
Show step-by-step solution
GivenU = − mB cos θ
AskedOrientation for maximum U
ConceptU is largest when cos θ is most negative.
FormulaU = − mB cos θ
Solutioncos θ is minimum (−1) at θ = 180°. Then U = + mB, the maximum value. θ = 180° means the moment is antiparallel to the field — the unstable position.
Answer: antiparallel to B
Q28Equilibrium
A current loop is in stable equilibrium in a uniform magnetic field when its magnetic moment is:
(a) parallel to B
(b) antiparallel to B
(c) perpendicular to B
(d) at 45° to B
Show step-by-step solution
GivenLoop free to rotate in a uniform field
AskedCondition for stable equilibrium
ConceptBoth θ = 0° and θ = 180° give zero torque, but only one of them restores the loop after a nudge.
Formulaτ = mB sin θ, U = − mB cos θ
SolutionAt θ = 0°: τ = 0 and U = −mB (minimum energy). A small nudge produces a torque that pushes it back — stable. At θ = 180°: τ = 0 but U = +mB (maximum energy); a nudge makes it flip over — unstable.
Answer: parallel to B
Q29Equilibrium
The work done in rotating a magnetic dipole from its stable equilibrium position to its unstable equilibrium position is:
(a) 0
(b) mB
(c) 2 mB
(d) 4 mB
Show step-by-step solution
GivenStable position θ = 0°, unstable position θ = 180°
AskedWork done W
ConceptRecognise the two named positions as 0° and 180°, then apply the work formula.
A coil of magnetic moment 10 A m² and moment of inertia 0.1 kg m² is free to rotate in a uniform field of 2 T. Starting from the aligned position, its angular speed after rotating through 90° is:
(a) 10 s⁻¹
(b) 20 s⁻¹
(c) 40 s⁻¹
(d) 100 s⁻¹
Show step-by-step solution
Givenm = 10 A m², B = 2 T, moment of inertia I = 0.1 kg m², rotation 0° → 90°
AskedAngular speed ω
ConceptCareful — here the coil starts aligned and the field does work ON it as it turns. Convert that work into rotational kinetic energy.
A current-carrying loop of irregular shape made of flexible wire is placed in a uniform magnetic field. The loop will:
(a) stay in its irregular shape
(b) collapse to a straight line
(c) become circular
(d) become square
Show step-by-step solution
GivenFlexible wire, irregular loop, uniform field
AskedFinal shape of the loop
ConceptThe loop tends towards maximum flux, which needs maximum area. For a fixed perimeter, a circle has the greatest area.
Formulam = IA; maximum flux at maximum area
SolutionThe forces on the loop push outward, tending to enlarge the enclosed area. The wire length (perimeter) is fixed. For a fixed perimeter, a circle encloses the maximum area — so the loop becomes circular, with its plane perpendicular to the field.
Answer: become circular
Q32Dipole · far field
The axial magnetic field of a small current loop of magnetic moment m at a large distance x is given by:
(a) μ₀m / 4πx³
(b) μ₀ (2m) / 4πx³
(c) μ₀m / 2πx²
(d) μ₀ (2m) / 4πx²
Show step-by-step solution
GivenSmall loop of moment m, axial point at distance x
AskedExpression for B
ConceptReplace IπR² by m in the simplified axial formula.
FormulaB = μ₀IA / 2πx³ with m = IA
SolutionB ≈ μ₀IR²/2x³. Multiply top and bottom by π: B = μ₀ I πR² / 2πx³ Since m = IπR², B = μ₀ m / 2πx³ Rewriting with 4π: B = (μ₀/4π)(2m/x³)
Answer: μ₀ (2m) / 4πx³
Q33Dipole · numeric
A small current loop of magnetic moment 10 A m² produces an axial field at a distance of 1 m from its centre of:
(a) 1 × 10⁻⁶ T
(b) 2 × 10⁻⁶ T
(c) 1 × 10⁻⁷ T
(d) 4 × 10⁻⁷ T
Show step-by-step solution
Givenm = 10 A m², x = 1 m
AskedAxial field B
ConceptStraight substitution into the dipole formula, using μ₀/4π = 10⁻⁷.
FormulaB = (μ₀/4π)(2m/x³)
SolutionB = 10⁻⁷ × (2 × 10) / (1)³ B = 10⁻⁷ × 20 = 2 × 10⁻⁶ T
Answer: 2 × 10⁻⁶ T
Q34Dipole · ratio
The axial field of a small current loop at a distance x is B. At a distance 2x it becomes:
(a) B / 2
(b) B / 4
(c) B / 8
(d) B / 16
Show step-by-step solution
GivenAxial field of a dipole at x and at 2x
AskedNew field value
ConceptUse the inverse-cube law. This is where students wrongly apply an inverse-square.
FormulaB ∝ 1 / x³
SolutionB′ / B = (x / 2x)³ = (1/2)³ = 1/8 B′ = B / 8
Answer: B / 8
Q35Dipole · analogy
The magnetic dipole formula can be obtained from the electric dipole formula by making the substitutions:
(a) μ₀ → ε₀ and m → p
(b) μ₀ → 1/ε₀ and m → p
(c) ε₀ → μ₀ and p → m
(d) 1/ε₀ → μ₀ and p → m
Show step-by-step solution
GivenAnalogy between electric and magnetic dipoles
AskedCorrect substitutions
ConceptNCERT states this analogy directly. Note the reciprocal on ε₀.
Formulaτ = p × E ↔ τ = m × B
SolutionThe electric result uses 1/4πε₀ where the magnetic result uses μ₀/4π. So μ₀ replaces 1/ε₀. And the magnetic moment m replaces the electric dipole moment p.
Answer: μ₀ → 1/ε₀ and m → p
Q36Dipole · orbiting charge
An electron moves in a circular orbit of radius r with speed v. The magnetic moment associated with this orbital motion is:
(a) evr
(b) evr / 2
(c) 2evr
(d) ev / 2r
Show step-by-step solution
GivenElectron of charge e, orbit radius r, speed v
AskedMagnetic moment m
ConceptAn orbiting charge is a current loop. Find the equivalent current first, then multiply by the area.
FormulaI = e/T = ev/2πr; m = IA
SolutionTime for one orbit: T = 2πr / v Equivalent current: I = e/T = ev / 2πr m = I A = (ev/2πr) × πr² m = evr / 2
Answer: evr / 2
Q37Dipole · orbiting charge
An electron (e = 1.6 × 10⁻¹⁹ C) revolves in an orbit of radius 0.53 × 10⁻¹⁰ m with a speed of 2.2 × 10⁶ m/s. Its orbital magnetic moment is about:
(a) 9.3 × 10⁻²⁴ A m²
(b) 9.3 × 10⁻²² A m²
(c) 1.9 × 10⁻²³ A m²
(d) 4.6 × 10⁻²⁴ A m²
Show step-by-step solution
Givene = 1.6 × 10⁻¹⁹ C, r = 0.53 × 10⁻¹⁰ m, v = 2.2 × 10⁶ m/s
AskedOrbital magnetic moment m
ConceptApply the result from the previous question. This value is the Bohr magneton.
A circular coil of 100 turns and radius 5 cm produces a magnetic field of 3.14 × 10⁻³ T at its centre. The current in the coil and its magnetic moment are respectively:
(a) 1.5 A, 1 A m²
(b) 2.5 A, 2 A m²
(c) 2.5 A, 4 A m²
(d) 5 A, 2 A m²
Show step-by-step solution
GivenN = 100, R = 5 cm = 0.05 m, B_centre = 3.14 × 10⁻³ T
AskedCurrent I and magnetic moment m
ConceptTwo sections in one question. Use the centre-of-coil formula to get I, then the moment formula.
FormulaB = μ₀NI/2R; m = N I πR²
SolutionFrom B = μ₀NI/2R: I = 2RB / (μ₀N) I = (2 × 0.05 × 3.14 × 10⁻³) / (4π × 10⁻⁷ × 100) I = 3.14 × 10⁻⁴ / (1.256 × 10⁻⁴) = 2.5 A m = N I πR² = 100 × 2.5 × 3.14 × (0.05)² = 250 × 3.14 × 2.5 × 10⁻³ m ≈ 1.96 ≈ 2 A m²
Answer: 2.5 A, 2 A m²
Q39Combined · 4.5 + 4.9
A circular coil of N turns and radius R carries a current I. If B is the field at its centre and m its magnetic moment, then the ratio B/m is:
(a) μ₀ / 2πR³
(b) μ₀ / 2R
(c) μ₀ / 2πR²
(d) 2πR³ / μ₀
Show step-by-step solution
GivenCoil of N turns, radius R, current I
AskedRatio B / m
ConceptWrite both expressions and divide. N and I cancel — a nice check that you have used the right pair of formulas.
FormulaB = μ₀NI/2R; m = N I πR²
SolutionB / m = (μ₀NI/2R) ÷ (N I πR²) N and I cancel from top and bottom. = μ₀ / (2R × πR²) = μ₀ / 2πR³
Answer: μ₀ / 2πR³
Q40Combined · 4.5 + 4.9
A wire of length L carrying current I is bent into a circular loop of one turn. If it is instead bent into n turns, then compared with the single turn, the field at the centre and the magnetic moment become respectively:
(a) n²B and m/n
(b) nB and nm
(c) B/n and nm
(d) n²B and n²m
Show step-by-step solution
GivenSame wire, 1 turn → n turns, same current
AskedNew field at centre and new moment
ConceptThe radius shrinks to R/n. Then track each formula separately — they move in opposite directions, which is the whole point of the question.
FormulaB = μ₀NI/2R; m = N I πR²
SolutionWire length fixed ⇒ new radius r = R/n. Field: B′ = μ₀ n I / 2(R/n) = μ₀ n² I / 2R = n²B Moment: m′ = n × I × π(R/n)² = IπR²/n = m/n
Answer: n²B and m/n
Q41Combined · galvanometer
In a moving coil galvanometer, the deflection θ of the coil is related to the current I through it by (k = torsional constant, N turns, area A, radial field B):
(a) θ = kI / NAB
(b) θ = NABI / k
(c) θ = NAB / kI
(d) θ = kNABI
Show step-by-step solution
GivenCoil of N turns, area A, in a radial field B, spring constant k
AskedRelation between θ and I
ConceptAt rest the magnetic torque is exactly balanced by the spring's restoring torque. The radial field keeps sin θ = 1 always.
FormulaNIAB = k θ
SolutionMagnetic torque on the coil = N I A B (radial field keeps the plane along B, so sin θ = 1). Restoring torque of the spring = k θ. At equilibrium: N I A B = k θ θ = N A B I / k, so θ ∝ I — a linear scale.
Answer: θ = NABI / k
Q42Combined · flux
A current loop free to rotate settles in its stable equilibrium position in a uniform external field. In this position the total magnetic flux through the loop is:
(a) zero
(b) minimum
(c) maximum
(d) unrelated to the orientation
Show step-by-step solution
GivenLoop in stable equilibrium
AskedTotal flux through the loop
ConceptIn stable equilibrium the area vector lines up with B, so the loop's own field adds to the external field.
FormulaΦ = B A cos θ, with θ = 0 in stable equilibrium
SolutionStable equilibrium ⇒ A is parallel to B ⇒ θ = 0. Φ = BA cos 0° = BA, the largest possible value. The loop's own field also points the same way, so the TOTAL flux is maximum.
Answer: maximum
Q43Torque · numeric
A rectangular coil of 100 turns and area 4 × 10⁻⁴ m² carries a current of 2 mA. It is placed in a uniform field of 0.5 T. The work done in rotating it through 180° from the stable position is:
(a) 20 μJ
(b) 40 μJ
(c) 80 μJ
(d) 160 μJ
Show step-by-step solution
GivenN = 100, A = 4 × 10⁻⁴ m², I = 2 mA = 2 × 10⁻³ A, B = 0.5 T
AskedWork done W for a 180° flip
ConceptFind the moment first, then use the standard 2mB result for a full flip.
Formulam = N I A; W = 2 m B
Solutionm = 100 × 2 × 10⁻³ × 4 × 10⁻⁴ = 8 × 10⁻⁵ A m² W = 2 m B = 2 × 8 × 10⁻⁵ × 0.5 W = 8 × 10⁻⁵ J = 80 μJ
Answer: 80 μJ
Q44Assertion–Reason
Assertion (A): A current-carrying loop placed in a uniform magnetic field experiences no net force but may still rotate. Reason (R): The forces on opposite arms of the loop are equal and opposite but do not act along the same line.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenLoop in a uniform field
AskedTruth of A and R, and whether R explains A
ConceptTest each statement separately, then ask whether the reason actually causes the assertion.
FormulaF_net = 0; τ = mB sin θ
SolutionA is TRUE: opposite forces cancel, so no net force; but a couple still acts, so it rotates. R is TRUE: the forces are equal and opposite, and they act along different lines. Does R explain A? Yes — forces on different lines are exactly what makes a couple and hence rotation without translation.
Answer: Both A and R are true, and R is the correct explanation of A
Q45Assertion–Reason
Assertion (A): A current loop is in stable equilibrium when its magnetic moment is antiparallel to the magnetic field. Reason (R): The torque on the loop is zero in that position.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is false but R is true
(d) Both A and R are false
Show step-by-step solution
GivenLoop with m antiparallel to B
AskedTruth of A and R
ConceptZero torque alone does not prove stability — check the energy as well. Both θ = 0° and θ = 180° give zero torque.
Formulaτ = mB sin θ; U = − mB cos θ
SolutionA is FALSE: antiparallel is the UNSTABLE position (U = +mB, maximum energy). Stable is parallel. R is TRUE: at θ = 180°, sin 180° = 0, so the torque is indeed zero. So A is false but R is true.
Answer: A is false but R is true
Q46Assertion–Reason
Assertion (A): The torque on a coil is zero when the plane of the coil is perpendicular to the magnetic field. Reason (R): In that position the angle between the magnetic moment and the field is 90°.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenPlane of the coil perpendicular to B
AskedTruth of A and R
ConceptThis tests the plane-versus-normal conversion directly — the single biggest trap in this topic.
Formulaτ = mB sin θ, θ measured from the normal
SolutionA is TRUE: plane ⊥ B ⇒ normal ∥ B ⇒ θ = 0° ⇒ sin θ = 0 ⇒ τ = 0. R is FALSE: in that position the angle between m and B is 0°, not 90°. So A is true but R is false.
Answer: A is true but R is false
Q47Match the following
Match the orientation of a current loop in a uniform field with the correct description:
(a) m parallel to B (b) m perpendicular to B (c) m antiparallel to B
(i) Maximum torque (ii) Stable equilibrium (iii) Unstable equilibrium
(a) a–ii, b–i, c–iii
(b) a–i, b–ii, c–iii
(c) a–iii, b–i, c–ii
(d) a–ii, b–iii, c–i
Show step-by-step solution
GivenThree orientations of a loop in a uniform field
AskedCorrect matching
ConceptWork out τ and U for each of the three angles, then match.
Formulaτ = mB sin θ; U = − mB cos θ
Solution(a) m ∥ B ⇒ θ = 0°: τ = 0, U = −mB (minimum) ⇒ stable equilibrium ⇒ (ii) (b) m ⊥ B ⇒ θ = 90°: τ = mB (maximum) ⇒ (i) (c) m antiparallel ⇒ θ = 180°: τ = 0, U = +mB (maximum) ⇒ unstable ⇒ (iii)
Answer: a–ii, b–i, c–iii
Q48Match the following
Match each quantity with its correct SI unit:
(a) Magnetic moment (b) Torque (c) Magnetic flux (d) Magnetic field
(i) tesla (ii) A m² (iii) weber (iv) N m
(a) a–ii, b–iv, c–iii, d–i
(b) a–i, b–iv, c–ii, d–iii
(c) a–ii, b–iii, c–iv, d–i
(d) a–iv, b–ii, c–i, d–iii
Show step-by-step solution
GivenFour quantities from this topic
AskedCorrect unit for each
ConceptReading the unit off each defining formula is the fastest route, and it also helps you spot the right formula in numerical questions.
Formulam = IA; τ = mB sin θ; Φ = BA
Solution(a) m = I × A ⇒ A m² ⇒ (ii) (b) Torque is force × distance ⇒ N m ⇒ (iv) (c) Flux is measured in weber ⇒ (iii) (d) Magnetic field is measured in tesla ⇒ (i)
Answer: a–ii, b–iv, c–iii, d–i
Q49Match the following
Match the wire configuration (same wire length L, same current I) with its magnetic moment:
(a) One circular turn (b) One square turn (c) n circular turns
(i) I L² / 16 (ii) I L² / 4πn (iii) I L² / 4π
(a) a–iii, b–i, c–ii
(b) a–i, b–iii, c–ii
(c) a–ii, b–i, c–iii
(d) a–iii, b–ii, c–i
Show step-by-step solution
GivenFixed wire length L, current I
AskedCorrect matching of moments
ConceptCompute the radius or side from the fixed perimeter in each case, then use m = NIA.
Formulam = N I A
Solution(a) Circle: 2πr = L ⇒ r = L/2π ⇒ A = L²/4π ⇒ m = IL²/4π ⇒ (iii) (b) Square: 4a = L ⇒ a = L/4 ⇒ A = L²/16 ⇒ m = IL²/16 ⇒ (i) (c) n turns: r = L/2πn ⇒ A = L²/4πn² ⇒ m = n I L²/4πn² = IL²/4πn ⇒ (ii)
Answer: a–iii, b–i, c–ii
Q50Concept · summary
Which of the following statements about a current loop in a UNIFORM magnetic field is INCORRECT?
(a) The net force on the loop is zero
(b) The torque depends on the area of the loop
(c) The torque depends on the shape of the loop for a given area
(d) The torque is zero when the plane of the loop is perpendicular to the field
Show step-by-step solution
GivenLoop in a uniform field
AskedThe incorrect statement
ConceptCheck each against τ = IAB sin θ. Shape appears nowhere in that formula.
Formulaτ = I A B sin θ; F_net = 0
Solution'Net force is zero' — correct in a uniform field. 'Torque depends on area' — correct, A appears in the formula. 'Torque depends on shape for a given area' — INCORRECT. Only the area enters, never the shape. 'Torque zero when plane ⊥ field' — correct, since then θ = 0.
Answer: The torque depends on the shape of the loop for a given area
Before the next ILTS — how to use this pack
Read Part 1 once, slowly. Do not skip to the questions. Every trap in Part 3 is explained there first.
Copy the formula sheet out by hand on a single side of paper. Writing it beats reading it.
Do Q1–Q12 (moment) in one sitting. These are the fastest marks in the topic.
Do Q13–Q26 (torque) the next day. Before each one, say out loud whether the question gives you the angle from the plane or from the normal.
Do Q27–Q36 (energy) and Q37–Q43 (dipole) together. These share one idea: U = −mB cos θ.
Q44–Q50 are the mixed ones — they combine this section with the coil field from 4.5. Save them for the final revision.
The ten-second drill. Take any twenty questions and, without calculating, say only which formula each one needs. If she gets 18 out of 20, the problem is arithmetic speed. If not, the problem is recognition — and that is what to fix first.
The three mistakes that cost the most marks in this topic
Measuring θ from the plane instead of the normal. "Plane parallel to B" means θ = 90°, maximum torque.
Forgetting N in m = NIA.
Not converting cm to m before squaring. 10 cm is 0.1 m, so the area carries a factor of 10⁻².