F = I L × B; magnitude F = B I L sin θ. For straight wire of length L in uniform B perpendicular to L: F = BIL.
-- NCERT Class 12 Physics, Ch. 4, p. 110Force on Current Conductor
8 MCQs2 revision cards9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 27 Sep 2026
Force on Current Conductor, explained for NEET
The wire formula has a sin θ in it and that is where the marks go. F = BIL sin θ — but θ is measured between the wire's own direction and B, not between the wire and anything else in the diagram. A wire lying along the field lines feels zero force, no matter how large the current. A wire lying across the field feels the maximum, BIL. Aspirants who memorise "F = BIL" and stop there lose the θ = 0 case entirely, and it is the case examiners like to set, because it looks like it should give a big answer.
NCERT Class 12 Physics Chapter 4 introduces this on printed page 124 as the vector product F = IL × B. Read the vector form once and the sin θ stops being a bolt-on: the cross product is zero when the two vectors are parallel, by construction.
Two more things the vector form tells you free of charge. First, L is a vector pointing along the current — so for a bent wire in a uniform field, only the straight-line displacement from entry point to exit point matters, not the path length. A semicircular wire of radius R behaves like a straight wire of length 2R. Second, the force direction is perpendicular to both L and B, which means it is perpendicular to the wire — a current-carrying wire in a magnetic field is pushed sideways, never dragged along itself.
The unit check is worth doing once: B in tesla, I in ampere, L in metre gives newton, because 1 T = 1 N·A⁻¹·m⁻¹. That identity is how tesla is defined, and it is a fast way to spot an option with the wrong power of ten.
Watch out: in numerical problems, resolve B into components parallel and perpendicular to the wire, or read the angle off the diagram carefully. An angle given "with the field" and an angle given "with the normal to the field" differ by 90°, and sin flips to cos.
Can you answer these Force on Current Conductor 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 expression F = BIL sin θ for the force on a straight current-carrying conductor in a uniform magnetic field, the angle θ is measured between
Show answer and why every option is right or wrong
Answer: B. B is correct. NCERT Class 12 Physics Chapter 4, printed page 124, gives the force as the vector product F = IL × B, where L points along the current; the angle in the magnitude of a cross product is the angle between the two vectors being multiplied, here the current direction and B.
Why A is wrong: A is wrong because the normal to a plane containing a single straight wire is not defined uniquely, and no normal appears in the vector product IL × B.
Why C is wrong: C is wrong because the force is always perpendicular to B (a property of the cross product), so that angle is fixed at 90° and could not appear as a variable θ.
Why D is wrong: D is wrong because the force is always perpendicular to the current direction as well, so that angle is also fixed at 90° and carries no information about the wire's orientation.
MCQ 2Easy RecallPractice
A straight wire carrying a steady current is placed in a uniform magnetic field so that the wire lies exactly along the field lines. The magnitude of the magnetic force on the wire is
Show answer and why every option is right or wrong
Answer: A. A is correct. With the wire along B, θ = 0 and sin 0 = 0, so F = BIL sin θ = 0 — the vector form F = IL × B on NCERT Class 12 Physics Chapter 4, printed page 124, gives zero for parallel vectors by construction.
Why B is wrong: B is wrong because BIL is the force when the wire is perpendicular to the field (θ = 90°, sin θ = 1), which is the opposite of the situation described.
Why C is wrong: C is wrong because BIL/2 would require sin θ = 0.5, i.e. θ = 30°, not θ = 0.
Why D is wrong: D is wrong because BIL sin θ can never exceed BIL — sin θ has a maximum value of 1 — so BIL√2 is not an attainable magnitude.
MCQ 3Easy RecallPractice
From the relation F = BIL sin θ, one tesla is equivalent to
Show answer and why every option is right or wrong
Answer: C. C is correct. Rearranging the force relation of NCERT Class 12 Physics Chapter 4, printed page 124, gives B = F/(IL sin θ), so the unit of B is newton divided by (ampere × metre), i.e. N·A⁻¹·m⁻¹.
Why A is wrong: A is wrong because it multiplies by ampere and metre instead of dividing, inverting the rearrangement of F = BIL sin θ.
Why B is wrong: B is wrong because it divides by ampere but multiplies by metre; the length must also be in the denominator.
Why D is wrong: D is wrong because it divides by metre but multiplies by ampere; the current must also be in the denominator.
MCQ 4Direct ApplicationPractice
A straight wire of length 0.250 m carries a steady current of 4.00 A and is placed perpendicular to a uniform magnetic field of magnitude 0.300 T. The magnitude of the force on the wire is
Show answer and why every option is right or wrong
Answer: C. C is correct. Perpendicular means θ = 90° and sin θ = 1, so F = BIL = (0.300)(4.00)(0.250) = 0.300 N, applying the relation on NCERT Class 12 Physics Chapter 4, printed page 124.
Why A is wrong: A is wrong because 3.00 × 10⁻² N is the product 0.300 × 4.00 × 0.250 mis-scaled by a factor of ten; check the decimal placement in 4.00 × 0.250 = 1.00.
Why B is wrong: B is wrong because 0.240 N results from using sin 53° or a similar non-unity sine factor; a perpendicular wire has sin θ = 1 exactly.
Why D is wrong: D is wrong because 1.20 N is B I L with the length taken as 1.00 m rather than 0.250 m — it is the force per metre, not the force on this wire.
MCQ 5Direct ApplicationPractice
A straight conductor of length 0.500 m carries a current of 6.00 A in a uniform magnetic field of 0.200 T. The conductor makes an angle of 30° (exact) with the field direction. The magnitude of the force on it is
Show answer and why every option is right or wrong
Answer: A. A is correct. F = BIL sin θ = (0.200)(6.00)(0.500)(sin 30°) = 0.600 × 0.500 = 0.300 N, following NCERT Class 12 Physics Chapter 4, printed page 124. The 30° is an exact angle, so sin 30° = 0.500 exactly and does not limit the significant figures.
Why B is wrong: B is wrong because 0.520 N uses cos 30° (≈ 0.866) instead of sin 30° — the angle given is with the field, so it goes into the sine, not the cosine.
Why C is wrong: C is wrong because 0.600 N omits the angular factor altogether, treating a 30° orientation as if it were perpendicular.
Why D is wrong: D is wrong because 1.20 N doubles the perpendicular value; there is no factor of 2 in F = BIL sin θ.
MCQ 6Concept TrapPractice
A straight wire carrying current I lies in a uniform magnetic field B. Which statement about the direction of the magnetic force on the wire is correct?
Show answer and why every option is right or wrong
Answer: D. D is correct. The force is the vector product IL × B given on NCERT Class 12 Physics Chapter 4, printed page 124, and a cross product is by definition perpendicular to both of its factors.
Why A is wrong: A is wrong because a force along the current direction would be parallel to L, which a cross product involving L can never be.
Why B is wrong: B is wrong because a force along B would be parallel to B, which the cross product IL × B can never be.
Why C is wrong: C is wrong because it states only half the condition — the force is indeed perpendicular to the wire, but it is simultaneously perpendicular to B, so the angle with B is fixed at 90° and not free.
MCQ 7CalculationPractice
A wire is bent into a semicircular arc of radius 0.100 m, and the two ends of the arc are the only points where it enters and leaves a region of uniform magnetic field 0.400 T directed perpendicular to the plane of the arc. The wire carries a current of 5.00 A. The magnitude of the net magnetic force on the semicircular portion is
Show answer and why every option is right or wrong
Answer: C. C is correct. In a uniform field, L in F = IL × B is the straight vector from entry point to exit point, so the effective length of the semicircle is its diameter, 2R = 0.200 m; then F = BIL = (0.400)(5.00)(0.200) = 0.400 N. The vector-product form on NCERT Class 12 Physics Chapter 4, printed page 124, is what licenses this replacement. The factor 2 in 2R is an exact counting factor and does not affect the significant figures.
Why A is wrong: A is wrong because 0.200 N uses the radius 0.100 m as the effective length instead of the end-to-end displacement 2R = 0.200 m.
Why B is wrong: B is wrong because 1.26 N uses the full circumference 2πR = 0.628 m as the effective length, as if the wire went all the way round.
Why D is wrong: D is wrong because 0.628 N uses the arc length πR = 0.314 m as the effective length; in a uniform field only the straight-line displacement between the ends counts, not the path travelled.
MCQ 8Direct ApplicationPractice
A straight wire of length 0.400 m carrying a current of 2.50 A is placed in a uniform magnetic field of magnitude 0.600 T. The wire is oriented at 60° (exact) to the field, and experiences a force F₁. The wire is then rotated within the same plane until it makes 30° (exact) with the field, giving a force F₂. The ratio F₁ : F₂ is
Show answer and why every option is right or wrong
Answer: D. D is correct. F ∝ sin θ, so F₁ : F₂ = sin 60° : sin 30° = (√3/2) : (1/2) = √3 : 1. The relation F = BIL sin θ is on NCERT Class 12 Physics Chapter 4, printed page 124; B, I and L are unchanged and cancel from the ratio.
Why A is wrong: A is wrong because 1 : √3 is the inverse ratio — it would follow from sin 30° : sin 60°, i.e. from swapping which orientation is F₁.
Why B is wrong: B is wrong because 1 : 2 comes from taking the ratio of the angles themselves (30 : 60 inverted) rather than the ratio of their sines.
Why C is wrong: C is wrong because 2 : 1 comes from the ratio of the angles 60 : 30, treating F as proportional to θ instead of to sin θ.
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Force on Current Conductor: quick recall before you leave
How do you solve a Force on Current Conductor question? A worked example
- 1
Given
• L = 0.750 m (length of wire in the field)• I = 8.00 A (steady current)• B = 0.250 T (uniform field)• θ = 45° (exact) — angle between the current direction and B
- 2
Required
Magnitude of the magnetic force F on the wire, and the orientation of that force relative to the wire.
- 3
Concept
A current-carrying conductor in a magnetic field experiences a force given by the vector product of its current-length vector with the field. Because it is a cross product, the magnitude carries a sine of the angle between the two vectors, and the direction is perpendicular to both. Here the wire is neither along nor across the field, so the sine factor is genuinely less than 1 and must be carried through.
- 4
Formula
F = IL × B, so the magnitude is F = B I L sin θ
(NCERT Class 12 Physics Chapter 4, printed page 124.) - 5
Substitution
F = (0.250 T)(8.00 A)(0.750 m)(sin 45°)
- 6
Calculation
B I L = 0.250 × 8.00 × 0.750 = 1.50 N
sin 45° = 1/√2 = 0.7071…
F = 1.50 × 0.7071 = 1.0606… N
The 45° is an exact problem-defined angle, so sin 45° = 1/√2 is a mathematical constant and contributes no significant-figure limit. The three significant figures come from B, I and L, each given to three. - 7
Final answer
F = 1.06 N, directed perpendicular to the wire (and simultaneously perpendicular to B), with the sense given by the right-hand rule applied to L × B.
- 8
Common trap
The 45° is given "with the field," so it goes straight into sin θ. If a stem instead gives the angle the wire makes with the normal to the field — or with a direction perpendicular to B — the angle with B is (90° − that angle), and sin becomes cos. Reading "45° to the field" as "45° from the perpendicular" happens to give the same number only at 45°; at any other angle it produces a wrong answer that still looks plausible. Always convert the stated angle into the angle between the current direction and B before substituting.
- 9
Similar NEET-style question
A straight wire of length 0.600 m carries 5.00 A in a uniform field of 0.400 T. The wire is oriented so that it makes an angle of 30° (exact) with a direction perpendicular to B, all in one plane. Find the force on the wire.
*(Answer: the angle with B is 90° − 30° = 60°, so F = 0.400 × 5.00 × 0.600 × sin 60° = 1.20 × 0.8660 = 1.04 N.)*
What to remember before solving Force on Current Conductor questions
Which Force on Current Conductor formulas do you need for NEET?
1 formula — click to collapse
Force on current-carrying wire
Force on straight current-carrying wire of length L in field B. Theta = angle between L and B.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | current | A |
| L | length | m |
| B | field | T |
| theta | angle | rad |
Valid when
- Uniform B
- Straight wire
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 10 formulas · 5 question patterns from its other lessons.
Force on Current Conductor questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys. — click to collapse
NEET 2023
1√5 IL
25 IL
3√3 IL
43 IL
NTA Answer: Option 2(final)
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
Sources
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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