Magnetic field due to current element I dl at distance r: dB = (μ₀/4π)(I dl × r̂)/r². μ₀ = 4π × 10⁻⁷ T·m/A.
-- NCERT Class 12 Physics, Ch. 4, p. 114Biot Savart Law
8 MCQs1 revision card9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismPYQ coverage: NEET 2021, 2022Official key: NTA-verifiedLast updated: 23 Sep 2026
Biot Savart Law, explained for NEET
The Biot-Savart law is not a field formula. It is the integrand that every field formula in this chapter is built from — and the commonest way to lose a mark here is to treat it as a ready-made answer for a whole wire.
NCERT Class 12 Physics, Chapter 4 (printed page 114) states it for one infinitesimal current element: dB = (μ₀/4π) · (I dl × r̂)/r². Read what each piece controls.
It is a vector cross product. The magnitude carries sin θ, where θ is the angle between the element dl and the position vector r pointing from the element to the field point. Directly along the wire's own axis, θ = 0, so that element contributes nothing there. Perpendicular to the element, sin θ = 1 and the contribution is maximal. The direction of dB is perpendicular to the plane containing dl and r.
It is an inverse-square law in r, like Coulomb's — but the resemblance stops there. Coulomb's source is a scalar charge; here the source is a vector element I dl, so dB has no radial component and the whole field curls around the current.
μ₀/4π = 10⁻⁷ T·m/A exactly, by the definition of μ₀ = 4π × 10⁻⁷ T·m/A. Carrying 10⁻⁷ instead of dividing 4π × 10⁻⁷ by 4π saves a step under time pressure.
The law applies to steady currents in free space. An isolated current element cannot physically exist on its own — current must flow in a closed circuit — so dB is a bookkeeping quantity. The measurable field is always the integral over the full circuit.
Where this bites in NEET: the long-straight-wire result B = μ₀I/(2πr) is what you get by integrating this law over an infinite wire. It is not what the law says. If a stem hands you a short element and a distance, the answer scales as I dl sin θ/r² — not as 1/r.
Can you answer these Biot Savart Law 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 Biot-Savart law dB = (μ₀/4π)·(I dl × r̂)/r², the unit vector r̂ points
Show answer and why every option is right or wrong
Answer: B. B is correct. In the statement of the law on printed page 114 of NCERT Class 12 Physics Chapter 4, r is the displacement vector drawn from the current element to the field point, so r̂ points source → field point.
Why A is wrong: A is wrong because dl already carries the current direction; r̂ is a separate vector, and if the two coincided the cross product would vanish everywhere.
Why C is wrong: C is wrong because reversing r̂ reverses the sign of the cross product and so reverses the predicted field direction — the convention is source to field point, not the reverse.
Why D is wrong: D is wrong because dB is perpendicular to r̂ by construction (it comes from dl × r̂), so r̂ can never lie along dB.
MCQ 2Direct ApplicationPractice
A short current element of length dl carries current I. A point P lies on the line obtained by extending the element along its own direction, at distance r from it. The magnitude of dB at P due to this element is
Show answer and why every option is right or wrong
Answer: C. C is correct. Along the element's own axis, the angle between dl and r̂ is 0°, so sin θ = 0 and the cross product vanishes — the sin θ dependence stated with the law on printed page 114 of NCERT Class 12 Physics Chapter 4.
Why A is wrong: A is wrong because it drops the sin θ factor entirely; that expression is the magnitude only at θ = 90°, not along the axis.
Why B is wrong: B is wrong because it uses 1/r rather than 1/r², and also ignores sin θ — the law is inverse-square in the distance from the element.
Why D is wrong: D is wrong because no factor of 2 appears in the Biot-Savart law for a single element; it also ignores sin θ = 0 at this position.
MCQ 3Concept TrapPractice
Which difference between the Biot-Savart law and Coulomb's law is correctly stated?
Show answer and why every option is right or wrong
Answer: B. B is correct. Chapter 4 of NCERT Class 12 Physics presents the law on printed page 114 with the cross product dl × r̂, which is perpendicular to r̂ — hence no radial component, unlike the radial Coulomb field of a point charge.
Why A is wrong: A is wrong because both laws carry 1/r² for their elementary sources; the inverse-cube behaviour appears only for a dipole's net field, not for the elementary law.
Why C is wrong: C is wrong because a charge can carry both fields at once — a moving charge in a region also containing an electric field feels both — so the statement about exclusivity is false.
Why D is wrong: D is wrong because it reverses the two: it is the Biot-Savart law that contains the cross product, while Coulomb's law is a scalar-source radial law.
MCQ 4Direct ApplicationPractice
A current element of length 2.00 × 10⁻³ m carries a current of 5.00 A. A point P is 0.100 m from the element, in a direction making 30.0° (exact) with the element. Taking μ₀/4π = 10⁻⁷ T·m/A (exact), the magnitude of dB at P is
Show answer and why every option is right or wrong
Answer: A. A is correct. dB = (μ₀/4π)·I dl sin θ/r² = 10⁻⁷ × (5.00 × 2.00 × 10⁻³ × sin 30.0°)/(0.100)² = 10⁻⁷ × (5.00 × 10⁻³)/(1.00 × 10⁻²) = 10⁻⁷ × 0.500 = 5.00 × 10⁻⁸ T, applying the law as stated on printed page 114 of NCERT Class 12 Physics Chapter 4.
Why B is wrong: B is wrong because it omits sin 30.0° = 0.500, which doubles the correct value.
Why C is wrong: C is wrong because it divides by r instead of r² — by 0.100 rather than by 1.00 × 10⁻² — which shrinks the answer by a factor of 10.
Why D is wrong: D is wrong because it uses cos 30.0° in place of sin 30.0° and also divides by r rather than r²; the angle in the law is measured between dl and r̂, and it is the sine of that angle that enters.
MCQ 5Easy RecallPractice
The SI unit of the quantity μ₀ appearing in the Biot-Savart law is
Show answer and why every option is right or wrong
Answer: B. B is correct. μ₀ = 4π × 10⁻⁷ T·m/A, the value quoted with the law in NCERT Class 12 Physics Chapter 4 (printed page 114).
Why A is wrong: A is wrong because ampere appears in the denominator, not the numerator — dimensionally, B = μ₀ × (current × length)/length², so μ₀ must divide out an ampere.
Why C is wrong: C is wrong because metre belongs in the numerator; T/(m·A) would make μ₀I dl/r² carry units of T/m², not T.
Why D is wrong: D is wrong because it inverts both the metre and the ampere relative to the correct T·m/A.
MCQ 6Concept TrapPractice
A student claims to have measured the magnetic field produced by a single isolated current element in the laboratory. The claim is unsound because
Show answer and why every option is right or wrong
Answer: C. C is correct. NCERT Class 12 Physics Chapter 4 (printed page 114) states the law for an element of a steady current, and steady current requires a closed circuit; the measurable field is always the integral of dB over the full circuit.
Why A is wrong: A is wrong because it inverts the condition of use: the law as stated applies to steady currents, and the displacement-current correction is what is needed for time-varying situations.
Why B is wrong: B is wrong because dB is non-zero at any point off the element's own axis; the objection is about physical realisability of the source, not about the value of dB.
Why D is wrong: D is wrong because the law as stated in the chapter is for free space, with μ₀ the permeability of free space.
MCQ 7CalculationPractice
For a fixed current element, dB is measured at a point P where the element-to-point direction makes 90.0° (exact) with dl, at distance r. A second point Q lies at distance 2r from the element, with the element-to-point direction making 30.0° (exact) with dl. The ratio dB at Q to dB at P is
Show answer and why every option is right or wrong
Answer: C. C is correct. dB ∝ sin θ/r². At P: sin 90.0°/r² = 1/r². At Q: sin 30.0°/(2r)² = 0.500/(4r²) = 1/(8r²). The ratio is therefore 1 : 8, following the law on printed page 114 of NCERT Class 12 Physics Chapter 4.
Why A is wrong: A is wrong because it accounts only for the sine factor (0.500) and ignores the fourfold reduction from doubling the distance in an inverse-square law.
Why B is wrong: B is wrong because it accounts only for the distance factor (1/4) and ignores sin 30.0° = 0.500.
Why D is wrong: D is wrong because it treats the distance dependence as 1/r³; the Biot-Savart law for a single element is inverse-square, and the inverse-cube behaviour belongs to a dipole's net field.
MCQ 8CalculationPractice
A current element I dl lies along the +x direction at the origin. Point P is on the +y axis at distance r. The direction of dB at P is
Show answer and why every option is right or wrong
Answer: D. D is correct. Here dl is along +x̂ and r̂ is along +ŷ, so dl × r̂ points along x̂ × ŷ = +ẑ — the cross-product direction required by the law on printed page 114 of NCERT Class 12 Physics Chapter 4.
Why A is wrong: A is wrong because dB is perpendicular to dl by the cross product, so it cannot lie along the element's own direction.
Why B is wrong: B is wrong because dB is perpendicular to r̂ as well, so it cannot point from the element towards P.
Why C is wrong: C is wrong because it is the direction of r̂ × dl, the cross product taken in the reverse order; the law fixes the order as dl × r̂.
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Biot Savart Law: quick recall before you leave
How do you solve a Biot Savart Law question? A worked example
Pattern: NEET pattern: biot savart long wire (topic_codes include PHY.U13.BIOT_SAVART_LAW), taken at the element level rather than as the integrated long-wire result.
- 1
Given
• Current element length dl = 1.50 × 10⁻³ m• Current I = 8.00 A• Distance to field point r = 5.00 × 10⁻² m• Angle between dl and r̂: θ = 60.0° (exact)• μ₀ = 4π × 10⁻⁷ T·m/A (exact), so μ₀/4π = 10⁻⁷ T·m/A (exact)
- 2
Required
The magnitude of the magnetic field dB contributed by this single element at the field point.
- 3
Concept
Each element of a current-carrying conductor contributes a field given by the Biot-Savart law. The contribution falls off as the inverse square of the distance and is weighted by the sine of the angle between the element and the line to the field point. This is the element-level contribution — the field of the complete conductor would require integrating over all such elements.
- 4
Formula
dB = (μ₀/4π) · (I dl sin θ)/r²
- 5
Substitution
dB = (10⁻⁷) × (8.00 × 1.50 × 10⁻³ × sin 60.0°)/(5.00 × 10⁻²)²
- 6
Calculation
Numerator: 8.00 × 1.50 × 10⁻³ = 1.20 × 10⁻² A·m
sin 60.0° = 0.866
1.20 × 10⁻² × 0.866 = 1.039 × 10⁻² A·m
Denominator: (5.00 × 10⁻²)² = 2.50 × 10⁻³ m²
dB = 10⁻⁷ × (1.039 × 10⁻²)/(2.50 × 10⁻³) = 10⁻⁷ × 4.157 = 4.157 × 10⁻⁷ T
Exact constants: μ₀/4π = 10⁻⁷ T·m/A is exact by the definition of μ₀, the angle 60.0° is a problem-defined exact value, and the 4π in μ₀ is a mathematical constant. None of these limits the significant-figure count. The measured inputs (8.00 A, 1.50 × 10⁻³ m, 5.00 × 10⁻² m) each carry three significant figures, so the answer carries three. - 7
Final answer
dB = 4.16 × 10⁻⁷ T, directed perpendicular to the plane containing dl and r̂.
- 8
Common trap
Reaching for B = μ₀I/(2πr) because the stem mentions a current and a distance. That expression is the result of integrating the Biot-Savart law over an infinite straight wire; it has no dl and no sin θ in it because both were consumed by the integration. When the stem gives you an element length and an angle, it is asking for the element-level law. Substituting into the long-wire formula here would give 3.20 × 10⁻⁵ T — larger by a factor of roughly 77, and dimensionally blind to the dl you were handed.
- 9
Similar NEET-style question
A current element of length 2.50 × 10⁻³ m carries 6.00 A. Find the magnitude of the field it contributes at a point 4.00 × 10⁻² m away, in a direction making 45.0° (exact) with the element. Take μ₀/4π = 10⁻⁷ T·m/A (exact). *(Answer: 6.63 × 10⁻⁷ T.)*
What to remember before solving Biot Savart Law questions
Which Biot Savart Law formulas do you need for NEET?
1 formula — click to collapse
Biot-Savart law
Magnetic field from current element. mu0 = 4*pi*10^-7 T*m/A.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | current | A |
| dl | element length | m |
| r | distance | m |
| mu0 | permeability | T*m/A |
Valid when
- Steady current
- Free space
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 10 formulas · 4 question patterns from its other lessons.
Biot Savart Law questions from past NEET papers
2 questions from NEET 2021, 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
1Statement I is incorrect and Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct and Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2021
14 Ia² and 3 Ia²
2√3 Ia² and 3 Ia²
33 Ia² and Ia²
43 Ia² and 4 Ia²
NTA Answer: Option 2(final)
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
How does NEET ask about Biot Savart Law?
1 recurring pattern from past papers — click to collapse
Field from long straight wire: B = mu_0 I/(2 pi r).
Direct ApplicationEasy
Common distractors
forgets pi in denominator
Confuses with circular loop formula
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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