The problems are on the Ampère's law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: The Biot–Savart law gives the field of a current element, dB = (μ₀/4π) I dl × r̂/r², and integrating it yields B = μ₀I/2πd for a long wire and μ₀I/2R at a loop's centre. These worked answers correct the 20 problems of the AP Physics C worksheet on Ampère's law, from direction of B to spot the error: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the law used, the substitution and the result with its unit in bold. Compare your choice of Amperian loop and your direction reasoning with the model, because a wrong enclosed current or a flipped direction is the commonest source of lost marks.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 12
Problem 1 Direction of B
1. The field lines are concentric circles around the wire.2. With the thumb out of the page the fingers curl anticlockwise, so B at a point on the right points up the page.
Problem 2 Long wire field
1. B = μ₀I/(2πd) = (2 × 10⁻⁷)(10)/0.050 = 4.0 × 10⁻⁵ T.2. B ∝ 1/d, so at 10 cm it is half, 20 μT.
Problem 3 Solenoid field
1. n = 500/0.25 = 2000 turns/m.2. B = μ₀nI = (4π × 10⁻⁷)(2000)(2.0) = 5.0 × 10⁻³ T.
Problem 4 Loop at its centre
1. B = μ₀NI/(2R) = (4π × 10⁻⁷)(20)(0.50)/(2 × 0.040) = 1.6 × 10⁻⁴ T.Problem 5 Force between wires
1. F/L = μ₀I₁I₂/(2πd) = (2 × 10⁻⁷)(25)/0.020 = 2.5 × 10⁻⁴ N/m.2. Opposite currents repel.
Problem 6 Reading a B–r graph
1. I = 2πrB/μ₀ = (0.020)(1.0 × 10⁻⁴)/(2 × 10⁻⁷) = 10 A.2. B ∝ 1/r, so at 10 cm B = 100 × 2.0/10 = 20 μT.
Problem 7 Ampère’s law statement
1. The line integral of B around a closed loop equals μ₀ times the net current through it.2. Inet = 3.0 − 1.0 = 2.0 A, so ∮B·dl = (4π × 10⁻⁷)(2.0) = 2.5 × 10⁻⁶ T·m.
Problem 8 Inside a thick wire
1. B = μ₀Ir/(2πR²) = (2 × 10⁻⁷)(5.0)(1.0 × 10⁻³)/(4.0 × 10⁻⁶) = 2.5 × 10⁻⁴ T.2. At the surface B = (2 × 10⁻⁷)(5.0)/(2.0 × 10⁻³) = 5.0 × 10⁻⁴ T.
3. B = (2 × 10⁻⁷)(5.0)/(4.0 × 10⁻³) = 2.5 × 10⁻⁴ T.
Problem 9 Finite wire segment
1. sin θ = 0.30/√(0.30² + 0.10²) = 0.30/0.316 = 0.949.2. B = (10⁻⁷)(4.0)(2 × 0.949)/0.10 = 7.6 × 10⁻⁶ T.
3. An infinite wire gives (2 × 10⁻⁷)(4.0)/0.10 = 8.0 μT, so the finite wire gives 95 % of it.
Problem 10 Loop on its axis
1. B = μ₀I/(2R) = (4π × 10⁻⁷)(3.0)/0.10 = 3.8 × 10⁻⁵ T.2. B = μ₀IR²/[2(x² + R²)3/2]: x² + R² = 0.0169, its 3/2 power is 2.197 × 10⁻³, so B = (3.77 × 10⁻⁶)(0.0025)/(2 × 2.197 × 10⁻³) = 2.1 × 10⁻⁶ T.
3. Ratio = 2.145/37.7 = 0.057.
Problem 11 Solenoid by Ampère
1. Ienc = nℓI = (3000)(0.10)(0.80) = 240 A (the sum of the turns’ currents).2. Bℓ = μ₀nℓI gives B = (4π × 10⁻⁷)(3000)(0.80) = 3.0 × 10⁻³ T.
3. Along bc and da the field is perpendicular to the path (or zero), so B · dl = 0.
Problem 12 Toroid
1. B = μ₀NI/(2πr) = (2 × 10⁻⁷)(400)(1.2)/0.10 = 9.6 × 10⁻⁴ T.2. B(0.080) = 1.2 × 10⁻³ T and B(0.12) = 8.0 × 10⁻⁴ T.
Problem 13 Opposite currents
1. At M the left wire’s field (current out) points down the page and the right wire’s field (current in) also points down, so they point the same way.2. Each B = (2 × 10⁻⁷)(6.0)/0.10 = 1.2 × 10⁻⁵ T, so the total is 2.4 × 10⁻⁵ T.
3. With the same direction the fields cancel and B = 0.
Problem 14 Force over a length
1. F = (F/L)L = (1.2 × 10⁻³)(2.5) = 3.0 × 10⁻³ N.2. The wires attract, so each force points toward the other wire.
Problem 15 A moving proton
1. B = (10⁻⁷)(1.6 × 10⁻¹⁹)(2.0 × 10⁶)/(1.0 × 10⁻²)² = 3.2 × 10⁻¹⁶ T.2. A 1 A current is about 6 × 10¹⁸ elementary charges passing every point each second, and such a wire gives 2 × 10⁻⁵ T at 1 cm, so one proton contributes very little.
Problem 16 Semicircular wire
1. On each straight section dl is parallel to r, so dl × r̂ = 0 at O.2. On the arc every dl is perpendicular to r, so B = (μ₀I/4π)(πR)/R² = μ₀I/(4R) = (4π × 10⁻⁷)(6.0)/(4 × 0.080) = 2.4 × 10⁻⁵ T.
Problem 17 Coaxial cable
1. B = μ₀I/(2πr) = (2 × 10⁻⁷)(2.0)/(3.0 × 10⁻³) = 1.3 × 10⁻⁴ T.2. The net enclosed current is 2.0 − 2.0 = 0, so B = 0 at 8.0 mm.
3. Outside the cable the opposite currents cancel, so there is no external field to disturb nearby signals.
Problem 18 Non-uniform current
1. I = ∫J 2πr dr = (2πJ₀/R)∫₀ᴿ r² dr = 2πJ₀R²/3 = 2π(4.0 × 10⁵)(9.0 × 10⁻⁶)/3 = 7.5 A.2. Ienc = 2πJ₀r³/(3R), so B = μ₀Ienc/(2πr) = μ₀J₀r²/(3R).
3. B(R) = μ₀J₀R/3 = (1.2566 × 10⁻⁶)(4.0 × 10⁵)(3.0 × 10⁻³)/3 = 5.0 × 10⁻⁴ T, and B(R/2) = B(R)/4 = 1.3 × 10⁻⁴ T.
Problem 19 Loop near a wire
1. F₁ = μ₀I₁I₂L/(2πd₁) = (2 × 10⁻⁷)(20)(5.0)(0.10)/0.050 = 4.0 × 10⁻⁵ N toward the wire.2. The far side is at 0.25 m and its current is opposite: F₂ = (2 × 10⁻⁷)(20)(5.0)(0.10)/0.25 = 8.0 × 10⁻⁶ N, away.
3. The net force is 4.0 × 10⁻⁵ − 0.8 × 10⁻⁵ = 3.2 × 10⁻⁵ N toward the wire. The side forces cancel.
Problem 20 Spot the error
1. B = μ₀I/(2πr) holds only outside, where all the current is enclosed. Inside, only the fraction r²/R² is enclosed.2. Emma: (2 × 10⁻⁷)(8.0)/(2.0 × 10⁻³) = 8.0 × 10⁻⁴ T.
3. Correct: B = μ₀Ir/(2πR²) = (2 × 10⁻⁷)(8.0)(2.0 × 10⁻³)/(1.6 × 10⁻⁵) = 2.0 × 10⁻⁴ T, a quarter of her value.
