Biot Savart's Law and Ampere's Circuital Law – Biot: NEET MCQs with Answers

Explanation: Use dB̄ = (μ₀I/4π)(dl̄ × r̂)/r², with r̄ drawn from the current element to O. The semicircle gives a field of magnitude μ₀I/4R, that is, (μ₀/4π)(πI/R). At the point (0, 0, R) the current runs along +Y and r̄ points along −Z, so dl̄ × r̄ points along −X, and B̄(arc) = −(μ₀/4π)(πI/R)î. Each straight part is a semi-infinite wire with O at perpendicular distance R from its end, so each gives a field of magnitude μ₀I/4πR. For the part at y = −R, the current is along −X and r̄ has a component along +Y, so the field is along −Z. For the part at y = +R, the current is along +X and r̄ has a component along −Y, so the field is again along −Z. Together they give −2(μ₀I/4πR)k̂. Adding, B̄ = −(μ₀/4π)(I/R)(πî + 2k̂).

Explanation: Applying the right-hand grip rule at any point on AB, the field due to one current points out of the page while that due to the other points into the page, and because the point is equidistant from both wires the magnitudes are equal — so they cancel. Along CD both fields point the same way and add, giving a non-zero resultant. Hence the field vanishes only on AB.

Explanation: The two arcs subtend 270° and 90° at P and carry currents i₁ and i₂. Since the wire is uniform, resistance is proportional to arc length, so R₁/R₂ = 270/90 = 3. Being in parallel, i₁R₁ = i₂R₂, giving i₂ = 3i₁. The fields are B₁ = (μ₀i₁/2R)(270/360) inward and B₂ = (μ₀i₂/2R)(90/360) outward. Substituting i₂ = 3i₁ makes the two contributions equal and opposite, so the resultant field at P is zero. (In general, the field at the centre of such a split loop is always zero.)

Explanation: Turn density n = 100/0.5 = 200 per metre, so B = μ₀nI = (4π × 10⁻⁷)(200)(2.5) = 6.28 × 10⁻⁴ T.

Explanation: Inside, B = μ₀Ir/2πa², so at r = a/2 it is μ₀I/4πa. Outside, B′ = μ₀I/2πr, so at r = 2a it is also μ₀I/4πa. The two are equal and the ratio is 1.

Explanation: Each wire produces μ₀I/2πd at P, and because the wires are mutually perpendicular the two field vectors at P are also perpendicular. Adding them at right angles gives B = (μ₀/2πd)√(I₁² + I₂²).

Explanation: The tesla is the SI unit of magnetic induction (magnetic flux density) B. One tesla is the induction at a point where a charge of one coulomb moving at right angles to the field with a speed of 1 m s⁻¹ experiences a force of 1 N, i.e. 1 T = 1 N A⁻¹ m⁻¹. Magnetic flux is measured in weber and magnetic moment in A m².

Explanation: Since B ∝ 1/r, increasing the distance from 4 cm to 12 cm — a factor of 3 — reduces the field to one-third of its value: B′ = 10⁻³/3 = 3.33 × 10⁻⁴ T.

Explanation: For a long straight wire, B = μ₀i/2πr, so B ∝ 1/r. Doubling the distance halves the field: B′ = 0.4/2 = 0.2 T.

Explanation: For points outside the wire, B = μ₀i/2πr depends only on the current and on the distance from the wire — not on its diameter. Since both the current and the distance are unchanged, the field far away remains the same.

Explanation: For an infinitely long straight wire, B = μ₀i/2πr, so r = μ₀i/2πB = (2 × 10⁻⁷ × 12)/(3 × 10⁻⁵) = 2.4 × 10⁻⁶/3 × 10⁻⁵ = 8 × 10⁻² m.

Explanation: The Biot–Savart law states dB ∝ i, dB ∝ dl, dB ∝ sin θ and dB ∝ 1/r². Combining these, dB = (μ₀/4π) i dl sin θ / r², whose vector form is dB = (μ₀/4π) i (dl × r)/r³ — the extra power of r in the denominator accounts for r being a vector of magnitude r rather than a unit vector.

Explanation: For fixed wire length L, the radius of an N-turn coil is r = L/2πN. Since B = μ₀NI/2r = μ₀N²I×π/L, we get B ∝ N². Therefore B₁/B₂ = 1²/2² = 1/4, i.e. 1 : 4.

Explanation: Applying Ampere's law to a circular loop drawn inside the hollow region, the enclosed current is zero, so B(2πr) = μ₀ × 0 and B = 0 inside. Outside, the whole current behaves as though concentrated on the axis, giving B = μ₀i/2πr. Hence the field exists only outside the pipe.

Explanation: B = μ₀NI/2r = (4π × 10⁻⁷ × 1000 × 0.1)/(2 × 0.1) = 2π × 10⁻⁴ = 6.28 × 10⁻⁴ T.