Magnetic force worksheet answers for AP Physics 2 (free PDF download)

Problem 1 Field of a cable

1. B = μ₀I/2πr = (4π × 10⁻⁷)(10) / (2π × 0.020) = 1.0 × 10⁻⁴ T.

2. B is proportional to 1/r, so doubling r halves the field to 5.0 × 10⁻⁵ T.

Problem 2 Field direction

1. The field lines are circles, directed counterclockwise seen from the front.

2. Right-hand grip rule: thumb along the current, fingers curl in the field direction.

Problem 3 Force on a proton

1. F = qvB = (1.6 × 10⁻¹⁹)(3.0 × 10⁵)(0.20) = 9.6 × 10⁻¹⁵ N.

2. The force is perpendicular to the velocity, so it does no work: the speed is constant.

Problem 4 Force on a wire

1. F = BIL = 0.50 × 4.0 × 0.25 = 0.50 N.

2. F is proportional to I, so it doubles to 1.0 N.

Problem 5 When is the force zero?

The force is zero when (1) the charge is at rest, (2) the velocity is parallel or antiparallel to the field (sin θ = 0), and (3) the particle is neutral (q = 0).

Problem 6 Which way?

1. Fingers right (v), curl into the page (B): the thumb points up the page.

2. The electron is negative, so the force is reversed: down the page.

Problem 7 Unit conversions

1. 250 mT = 0.25 T.

2. 45 μT = 4.5 × 10⁻⁵ T.

3. From F = BIL, 1 T = 1 N/(A·m).

Problem 8 Matching the Earth’s field

r = μ₀I / 2πB = (4π × 10⁻⁷)(30) / (2π × 5.0 × 10⁻⁵) = 0.12 m, so 12 cm.

Problem 9 Proton in a field

1. r = mv/(qB) = (1.67 × 10⁻²⁷)(4.0 × 10⁵) / [(1.6 × 10⁻¹⁹)(0.50)] = 8.4 mm.

2. T = 2πm/(qB) = 2π(1.67 × 10⁻²⁷) / (8.0 × 10⁻²⁰) = 1.3 × 10⁻⁷ s.

Problem 10 Wire at an angle

1. F = BIL sin θ = 0.25 × 8.0 × 0.60 × sin 30° = 0.60 N.

2. The greatest force occurs at 90°.

Problem 11 Velocity selector

1. The electric and magnetic forces cancel: qE = qvB.

2. v = E/B = 4.0 × 10³ / 0.020 = 2.0 × 10⁵ m/s.

Problem 12 Two parallel wires

1. F/L = μ₀I₁I₂ / 2πd = (4π × 10⁻⁷)(5.0)(5.0) / (2π × 0.020) = 2.5 × 10⁻⁴ N/m.

2. Parallel currents attract.

Problem 13 Radius and field data

1. r is inversely proportional to B: doubling B halves r.

2. r × B = 11.4 × 10⁻³ × 1.0 × 10⁻³ = 1.14 × 10⁻⁵ T·m; the other rows give 1.14 × 10⁻⁵ and 1.14 × 10⁻⁵ T·m. It is constant.

3. r = mv/(eB), so v = e(rB)/m = (1.6 × 10⁻¹⁹)(1.14 × 10⁻⁵) / (9.11 × 10⁻³¹) = 2.0 × 10⁶ m/s.

Problem 14 Accelerating a bar

1. F = BIL = 0.40 × 12 × 0.15 = 0.72 N.

2. a = F/m = 0.72 / 0.050 = 14 m/s².

Problem 15 Electron versus proton

1. r = mv/(qB), and the charge magnitudes are equal, so r is proportional to mass: r_p / r_e = m_p / m_e = 1.67 × 10⁻²⁷ / 9.11 × 10⁻³¹ = 1.8 × 10³.

2. The charges are opposite, so they circulate in opposite directions.

Problem 16 Field between two wires

1. Each wire gives B = (4π × 10⁻⁷)(6.0) / (2π × 0.050) = 2.4 × 10⁻⁵ T at P. For opposite currents the two fields point the same way at P, so the total is 4.8 × 10⁻⁵ T.

2. For currents in the same direction the two fields are equal and opposite at P, so the total is zero.

Problem 17 Levitating a wire

1. mg = BIL, so I = mg / (BL) = (0.012)(9.81) / (0.060 × 0.50) = 3.9 A.

2. With B to the right and an upward force, the right-hand rule needs the current out of the page, as drawn.

Problem 18 Mass spectrometer

1. qV = ½mv², so v = √(2qV/m) = √[2(1.6 × 10⁻¹⁹)(500) / 6.5 × 10⁻²⁶] = 5.0 × 10⁴ m/s.

2. r = mv/(qB) = (6.5 × 10⁻²⁶)(4.96 × 10⁴) / [(1.6 × 10⁻¹⁹)(0.100)] = 0.20 m.

Problem 19 Spiral path

1. F = qvB sin θ = (1.6 × 10⁻¹⁹)(5.0 × 10⁵)(0.30) sin 60° = 2.1 × 10⁻¹⁴ N.

2. v⊥ = 5.0 × 10⁵ sin 60° = 4.33 × 10⁵ m/s. r = mv⊥/(qB) = (1.67 × 10⁻²⁷)(4.33 × 10⁵) / (1.6 × 10⁻¹⁹ × 0.30) = 1.5 cm.

3. T = 2πm/(qB) = 2.19 × 10⁻⁷ s. v∥ = 5.0 × 10⁵ cos 60° = 2.5 × 10⁵ m/s. Distance = v∥T = 5.5 cm.

Problem 20 Spot the errors

1. The magnetic force is always perpendicular to the velocity, so it changes the direction but does no work. The speed and kinetic energy are unchanged.

2. F = BIL sin θ, and sin 0° = 0, so a wire parallel to the field feels no force. F = 0.20 × 3.0 × 0.50 × 0 = 0 N.