Formula
Torque on current loop
τ = N B I A sin θ, where N is number of turns, A is loop area, θ is angle between B and normal to loop. Magnetic dipole moment m = NIA; τ = m × B.
-- NCERT Class 12 Physics, Ch. 4, p. 126Torque Current Loop
8 MCQs1 revision card9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Torque Current Loop, explained for NEET
The angle in τ = NIAB sin θ is not the angle you can see. A stem that reads "the plane of the coil makes 30° with the field" has not handed you θ. In this expression the angle is measured between the field and the normal to the loop — the direction of the loop's magnetic moment, m = NIA. If the plane makes α with B, then θ = 90° − α. Substituting sin 30° where sin 60° belongs changes the answer by a factor of about 1.7, and both numbers usually sit on the option list.
The result is stated in NCERT Class 12 Physics, Chapter 4, page 133: a coil of N turns and area A carrying steady current I in a uniform field B experiences a torque τ = NIAB sin θ, written compactly as τ = m × B with m = NIA along the normal.
Where it comes from is worth holding onto. In a uniform field the forces on opposite sides of the loop are equal and opposite, so the net force on the loop is zero. What survives is a couple: a turning effect with no translation. Two orientations serve as anchors — when m lies along B (θ = 0) the torque vanishes; when the plane of the loop contains B (θ = 90°) the torque is maximum at NIAB.
Two conditions travel with the formula as stated: the field must be uniform, and the loop planar. Drop either and the single-line expression no longer applies.
For NEET this is a medium-weight item within its chapter — it surfaces in roughly one paper in two, most often as recall of the angle convention or a one-step substitution, with medium negative-marking risk.
Watch out: before you compute, decide whether the stem gave you the angle to the plane or to the normal, and write θ down explicitly.
Can you answer these Torque Current Loop MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the torque expression τ = NIAB sin θ for a planar current loop in a uniform magnetic field, the angle θ is measured between which pair of directions?
Show answer and why every option is right or wrong
Answer: C. C is correct. The torque is τ = m × B, and m points along the normal to the loop's plane, so θ is the angle between that normal and B, as stated in NCERT Class 12 Physics, Chapter 4, page 126.
Why A is wrong: A is wrong because the angle to the plane is the complement of θ; using it directly replaces sin θ with cos θ and is the single most common substitution error in this topic.
Why B is wrong: B is wrong because the current direction varies from side to side around the loop, so it cannot define one angle for the whole loop; the normal is the only direction the loop as a whole defines.
Why D is wrong: D is wrong because the torque expression depends on the loop's area and orientation, not on which side is longer; a circular loop has no sides at all yet obeys the same formula.
MCQ 2Easy RecallPractice
A rigid closed current loop is placed in a uniform magnetic field. Which statement about the net force and the net torque on the loop is correct?
Show answer and why every option is right or wrong
Answer: A. A is correct. In a uniform field the forces on opposite sides of the loop are equal and opposite, so they cancel as a net force but constitute a couple, giving a torque τ = NIAB sin θ (NCERT Class 12 Physics, Chapter 4, page 125).
Why B is wrong: B is wrong because it inverts the actual result: the cancellation applies to the force, and the surviving couple is precisely what makes the torque non-zero.
Why C is wrong: C is wrong because it would leave the loop with no turning effect at all, contradicting the torque expression stated for this configuration.
Why D is wrong: D is wrong on both counts: the net force is zero in a uniform field, and the torque itself vanishes at the orientations where the normal lies along B.
MCQ 3Easy RecallPractice
The expression τ = NIAB sin θ for the torque on a current-carrying coil is stated under which conditions?
Show answer and why every option is right or wrong
Answer: D. D is correct. The two conditions attached to this result are a uniform magnetic field and a loop lying in a plane, as set out with the formula in NCERT Class 12 Physics, Chapter 4, page 133.
Why A is wrong: A is wrong because a non-uniform field leaves the side forces unequal, so a net force appears and the simple couple result no longer describes the loop.
Why B is wrong: B is wrong because a non-planar loop has no single normal direction, and without one normal the angle θ in the expression is undefined.
Why C is wrong: C is wrong because a radial field is a special engineered arrangement, not a condition of the general result; the general statement requires a uniform field.
MCQ 4Direct ApplicationPractice
A coil of 100 turns (exact) and area 2.0 × 10⁻³ m² carries a steady current of 3.0 A. It is placed in a uniform magnetic field of 0.50 T with its normal at 30° to the field direction. The magnitude of the torque on the coil is
Show answer and why every option is right or wrong
Answer: B. B is correct. τ = NIAB sin θ = 100 × 3.0 × 2.0 × 10⁻³ × 0.50 × sin 30° = 0.30 × 0.500 = 1.5 × 10⁻¹ N·m, using the formula on NCERT Class 12 Physics, Chapter 4, page 125.
Why A is wrong: A is wrong because it uses sin 60° — the result of treating the given 30° as the angle to the plane rather than to the normal, which the stem states explicitly.
Why C is wrong: C is wrong because it drops sin θ entirely, which is the value NIAB takes only at θ = 90°, not at the stated 30°.
Why D is wrong: D is wrong because it omits the factor N = 100; the torque on an N-turn coil is N times that on a single turn.
MCQ 5Direct ApplicationPractice
A planar current loop can be freely oriented in a uniform magnetic field. The torque on it has its maximum magnitude when
Show answer and why every option is right or wrong
Answer: A. A is correct. If the plane contains B, the normal is perpendicular to B, so θ = 90°, sin θ = 1 and τ reaches its largest value NIAB (NCERT Class 12 Physics, Chapter 4, page 125).
Why B is wrong: B is wrong because the moment along the field means θ = 0, where sin θ = 0 and the torque vanishes — this is the minimum, not the maximum.
Why C is wrong: C is wrong because sin 45.0° ≈ 0.707, which is smaller than the value 1 reached at θ = 90°.
Why D is wrong: D is wrong because the torque carries an explicit sin θ factor, so it varies between 0 and NIAB as the loop is turned.
MCQ 6Direct ApplicationPractice
A coil of 50 turns (exact) and area 4.0 × 10⁻³ m² carries a current of 2.0 A in a uniform magnetic field of 0.30 T. The field is directed perpendicular to the plane of the coil. The magnitude of the torque on the coil is
Show answer and why every option is right or wrong
Answer: D. D is correct. A field perpendicular to the plane is parallel to the normal, so θ = 0, sin θ = 0 and τ = 0; the coil is in equilibrium rather than being turned (NCERT Class 12 Physics, Chapter 4, page 126).
Why A is wrong: A is wrong because it is NIAB with sin θ = 1, obtained by feeding the word 'perpendicular' straight into the formula as θ = 90° instead of resolving it to the normal.
Why B is wrong: B is wrong because it repeats that same angle slip and also drops the factor N = 50.
Why C is wrong: C is wrong because it carries a decimal slip in the area, using 4.0 × 10⁻² m² in place of the stated 4.0 × 10⁻³ m².
MCQ 7CalculationPractice
A coil of 20 turns (exact) and area 5.0 × 10⁻² m² carries a current of 4.0 A. It lies in a uniform magnetic field of 0.10 T with the plane of the coil making an angle of 30° with the field direction. The magnitude of the torque on the coil is
Show answer and why every option is right or wrong
Answer: C. C is correct. The plane is at 30° to B, so the normal is at θ = 90° − 30° = 60°; then τ = 20 × 4.0 × 5.0 × 10⁻² × 0.10 × sin 60° = 0.40 × 0.866 = 3.5 × 10⁻¹ N·m (NCERT Class 12 Physics, Chapter 4, page 125).
Why A is wrong: A is wrong because it substitutes sin 30° — using the given plane angle as θ without converting to the normal angle.
Why B is wrong: B is wrong because it uses sin θ = 1, which applies only when the plane contains the field, not when it makes 30° with it.
Why D is wrong: D is wrong because the angle is handled correctly but the factor N = 20 has been dropped, giving the single-turn torque.
MCQ 8CalculationPractice
A planar coil in a uniform magnetic field experiences a torque of 1.20 × 10⁻² N·m when its plane makes an angle of 60.0° with the field. With the current, the field and the coil unchanged, the coil is turned until its plane makes 30.0° with the field. The new torque is
Show answer and why every option is right or wrong
Answer: B. B is correct. A plane angle of 60.0° gives θ = 30.0°, so NIAB = 1.20 × 10⁻² / sin 30.0° = 2.40 × 10⁻² N·m; at a plane angle of 30.0°, θ = 60.0° and τ = 2.40 × 10⁻² × sin 60.0° = 2.08 × 10⁻² N·m (NCERT Class 12 Physics, Chapter 4, page 125).
Why A is wrong: A is wrong because it simply halves the original torque, as though τ scaled with the plane angle itself rather than with the sine of the normal angle.
Why C is wrong: C is wrong because it scales by sin 30.0°/sin 60.0°, applying the sine to the plane angles directly instead of converting each to its normal angle first.
Why D is wrong: D is wrong because 2.40 × 10⁻² N·m is the maximum torque NIAB, reached only when the plane contains the field, not at a plane angle of 30.0°.
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Torque Current Loop: quick recall before you leave
How do you solve a Torque Current Loop question? A worked example
- 1
Given
• Number of turns, N = 250 (exact, a counting integer)• Radius of the circular coil, r = 4.00 × 10⁻² m• Current, I = 1.50 A• Uniform magnetic field, B = 0.200 T• Angle between the normal to the coil and the field, θ = 60.0°
- 2
Required
The magnitude of the torque acting on the coil.
- 3
Concept
In a uniform field the forces on opposite parts of the loop cancel as a net force but form a couple. The turning effect is τ = m × B, where the magnetic moment m = NIA points along the normal to the coil's plane. The relevant angle is the one between that normal and B — here given directly as 60.0°.
- 4
Formula
τ = NIAB sin θ, with A = πr² for a circular coil.
- 5
Substitution
A = π(4.00 × 10⁻² m)² = π × 1.600 × 10⁻³ m²
τ = 250 × 1.50 A × A × 0.200 T × sin 60.0° - 6
Calculation
A = 5.027 × 10⁻³ m²
m = NIA = 250 × 1.50 × 5.027 × 10⁻³ = 1.885 A·m²
τ = 1.885 × 0.200 × sin 60.0° = 0.3770 × 0.8660 = 0.3265 N·m
The number of turns (250) is a counting integer and π is a mathematical constant; neither is a measurement, so neither contributes to the significant-figure count. The three significant figures come from 4.00 × 10⁻² m, 1.50 A and 0.200 T. - 7
Final answer
τ = 3.26 × 10⁻¹ N·m
- 8
Common trap
Had the stem instead said "the plane of the coil makes 60.0° with the field", θ would be 30.0°, not 60.0°, and the answer would be 1.885 × 0.200 × 0.500 = 1.88 × 10⁻¹ N·m. Both numbers are plausible-looking, so the deciding step is reading which angle the stem supplies. Write θ down explicitly before substituting.
- 9
Similar NEET-style question
The same coil is now turned until its plane contains the field direction, with everything else unchanged. By what factor does the torque change, and what is its new value? (Answer: the normal is now at 90.0° to B, so τ = NIAB = 3.77 × 10⁻¹ N·m, a factor of 1/sin 60.0° ≈ 1.15 larger.)
What to remember before solving Torque Current Loop questions
Which Torque Current Loop formulas do you need for NEET?
1 formula — click to collapse
Torque on current loop
Torque on N-turn loop of area A carrying I in field B; magnetic moment m = NIA.
| Symbol | Quantity | SI Unit |
|---|---|---|
| N | turns | - |
| A | loop area | m^2 |
| B | field | T |
| m | magnetic moment | A*m^2 |
| theta | angle between m and B | rad |
Valid when
- Uniform B
- Loop in plane
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 10 formulas · 5 question patterns from its other lessons.
Torque Current Loop questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
Sources
NCERT refs: Class 12 Physics Chapter 4, p.133
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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