The problems are on the magnetic fields worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A magnetic field exerts a force on moving charges and on current-carrying wires. These worked answers correct the 20 problems of the Physics 11-12 worksheet on magnetic fields, from force on a lab wire to checking a student’s solution: each solution shows the formula, the substitution and the final answer with its unit.
Check your method as well as the final number: each answer shows the formula, the substitution and the result with its unit. Directions come from the left-hand rule for conventional current. Small rounding differences in the last digit are acceptable.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB D.3
Problem 2 Direction of the force
1. Current north, field down: the force is perpendicular to both, so it is horizontal. By the left-hand rule (or the cross product I × B) it points west.2. Reversing the current reverses the force: east.
Problem 3 Electron in a field
F = Bqv = 0.20 × 1.60 × 10⁻¹⁹ × 5.0 × 10⁶ = 1.6 × 10⁻¹³ N.Problem 4 Finding the field
B = F / (IL) = 0.054 / (6.0 × 0.15) = 0.054 / 0.90 = 0.060 T.Problem 5 Wire at an angle
1. F = BIL sin θ = 0.25 × 2.0 × 0.40 × sin 30° = 0.20 × 0.50 = 0.10 N.2. sin 0° = 0, so the force is zero.
Problem 6 Power cable and Earth’s field
1. 45 µT = 45 × 10⁻⁶ T = 4.5 × 10⁻⁵ T.2. F = BIL = 4.5 × 10⁻⁵ × 250 × 80 = 0.90 N, small for a cable of that length.
Problem 7 Speed unchanged
1. The magnetic force is always perpendicular to the velocity, so it has no component along the motion and does no work.2. Because no work is done, the kinetic energy is constant; only the direction of the velocity changes.
Problem 8 Proton orbit radius
r = mv/(qB) = (1.67 × 10⁻²⁷ × 4.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.80) = 6.68 × 10⁻²¹ / 1.28 × 10⁻¹⁹ = 5.2 cm.Problem 9 Cyclotron frequency of an electron
1. T = 2πm/(qB) = 2π × 9.11 × 10⁻³¹ / (1.60 × 10⁻¹⁹ × 1.5 × 10⁻³) = 2.4 × 10⁻⁸ s.2. f = 1/T = 4.2 × 10⁷ Hz.
3. Neither changes: the period does not depend on speed, because the radius doubles with the speed.
Problem 10 Selecting a speed
1. v = E/B = 2.4 × 10⁴ / 0.080 = 3.0 × 10⁵ m/s.2. The magnetic force qvB grows with v while qE stays the same, so the magnetic force wins and the faster particle is deflected out of the straight line, towards the magnetic-force side.
Problem 11 Electron gun then field
1. eV = ½mv², so v = √(2eV/m) = √(2 × 1.60 × 10⁻¹⁹ × 150 / 9.11 × 10⁻³¹) = 7.3 × 10⁶ m/s.2. r = mv/(qB) = 9.11 × 10⁻³¹ × 7.3 × 10⁶ / (1.60 × 10⁻¹⁹ × 1.0 × 10⁻³) = 4.1 cm.
Problem 12 Floating wire
1. The upward magnetic force must equal the weight: BIL = mg, so BI = (m/L)g.2. I = (m/L)g / B = 4.0 × 10⁻³ × 9.81 / 0.050 = 0.78 A.
Problem 13 Reading a force graph
1. The line passes through (5.0 A, 0.100 N): gradient = 0.100 / 5.0 = 0.020 N/A.2. F = (BL) I, so gradient = BL and B = 0.020 / 0.25 = 0.080 T.
3. With no current there is no force on the wire, so F = 0 when I = 0.
Problem 14 Proton against alpha particle
1. r = mv/(qB), so r(α)/r(p) = (m(α)/m(p)) × (q(p)/q(α)) = (6.64 × 10⁻²⁷ / 1.67 × 10⁻²⁷) × (1/2) = 3.98 × 0.5 = 1.99 (about 2).2. The alpha is four times heavier, which widens the circle, but it carries twice the charge, which tightens it. The net effect is a circle twice as large.
Problem 15 Tracks in a bubble chamber
1. For a positive charge moving right with B into the page, F = qv × B points up, so the track curves upwards: track 1 is positive, track 2 is negative.2. p = mv = qBr. With equal |q| and the same B, the larger radius means larger momentum: track 1.
3. The force is always perpendicular to the velocity and constant in size, so it acts as a centripetal force.
Problem 16 Proton cyclotron
1. v = qBr/m = 1.60 × 10⁻¹⁹ × 1.2 × 0.45 / 1.67 × 10⁻²⁷ = 5.17 × 10⁷ m/s.2. KE = ½mv² = 0.5 × 1.67 × 10⁻²⁷ × (5.17 × 10⁷)² = 2.24 × 10⁻¹² J.
3. 2.24 × 10⁻¹² / 1.60 × 10⁻¹⁹ = 1.40 × 10⁷ eV = 14.0 MeV.
Problem 17 Spiral path
1. v⊥ = v sin 60° = 3.46 × 10⁶ m/s, v∥ = v cos 60° = 2.00 × 10⁶ m/s.2. r = mv⊥/(eB) = 9.11 × 10⁻³¹ × 3.46 × 10⁶ / (1.60 × 10⁻¹⁹ × 2.0 × 10⁻³) = 0.99 cm.
3. T = 2πm/(eB) = 1.79 × 10⁻⁸ s, so pitch = v∥T = 2.00 × 10⁶ × 1.79 × 10⁻⁸ = 3.6 cm.
Problem 18 Coil in a field
1. F = N B I L = 40 × 0.15 × 0.80 × 0.050 = 0.24 N.2. The current flows the opposite way in the other side, so its force is equal and opposite.
3. Couple = F × perpendicular distance = 0.24 × 0.030 = 7.2 × 10⁻³ N m.
Problem 19 Separating chlorine ions
1. qV = ½mv² gives v = √(2qV/m); substituting into r = mv/(qB) gives r = (1/B)√(2mV/q).2. For 35 u: m = 5.81 × 10⁻²⁶ kg, r = 7.62 cm. For 37 u: m = 6.14 × 10⁻²⁶ kg, r = 7.84 cm.
3. The landing points are 2r from the entry slit, so the separation is 2(r₃₇ − r₃₅) = 4.3 mm.
Problem 20 Checking a student’s solution
1. r = 1.67 × 10⁻²⁷ × 2.0 × 10⁶ / (1.60 × 10⁻¹⁹ × 0.40) = 3.34 × 10⁻²¹ / 6.4 × 10⁻²⁰ = 5.2 cm = 0.052 m, ten times smaller than claimed.2. T = 2πr/v = 2π × 0.0522 / 2.0 × 10⁶ = 1.6 × 10⁻⁷ s.
3. Estimate 1.7 × 2 ÷ (1.6 × 0.4) ≈ 5, and 10⁻²⁷⁺⁶ ÷ 10⁻¹⁹ = 10⁻², so r should be about 5 × 10⁻² m, a few centimetres.
