Physics

1. 1.

   A straight wire of length 0.150 m carries 4.00 A perpendicular to a uniform 0.200 T field. Calculate magnetic force and state its direction relative to current and field.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Only the component of wire length perpendicular to the field contributes; here the entire length does.
   2. No unique compass direction can be stated without specifying current and field orientations.

   Marking points

   • F = BIL sin(90 degrees).
   • F = 0.120 N.
   • Force is perpendicular to both current and field, with direction given by the motor rule.

   Examiner tip: Do not invent a compass direction from magnitudes alone.

2. 2.

   A flat loop of area 0.0200 m^2 is in a 0.300 T uniform field. Its normal is 60.0 degrees to the field. Find flux through the loop.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Project the area perpendicular to the field using the angle of its normal.
   2. The loop is not a multi-turn coil, so no turn multiplier is needed.

   Marking points

   • Flux Phi = BA cos theta, where theta is the normal-field angle.
   • Substitution: 0.300*0.0200*cos(60 degrees).
   • Phi = 0.00300 Wb.

   Examiner tip: If the angle were to the plane instead, the trigonometric factor would change.

3. 3.

   A proton moves at 2.00e6 m s^-1 perpendicular to a 0.500 T field in vacuum. Find its circular radius and explain why speed stays constant. Use m = 1.67e-27 kg and q = 1.60e-19 C; neglect relativistic effects.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The magnetic force supplies the radial resultant, changing direction but not kinetic energy.
   2. Cancel one v when equating the force expressions; the radius is proportional to speed, not speed squared.

   Marking points

   • qvB = mv^2/r, so r = mv/(qB).
   • r = 0.0418 m (3 significant figures).
   • Magnetic force is perpendicular to velocity and does no work.

   Examiner tip: A magnetic field bends the path; it does not accelerate the particle along its path.

4. 4.

   Flux through each turn of a 200-turn coil decreases uniformly from 0.00600 Wb to 0.00100 Wb in 0.0500 s. Find induced emf magnitude. Explain Lenz's law for the induced current if the circuit is closed.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The given flux is per turn, so multiply its rate of change by the turn count.
   2. Lenz's law opposes the change, not necessarily the original field; no clockwise direction is determinable without a viewing orientation.

   Marking points

   • Magnitude emf = N|delta Phi|/delta t.
   • Induced emf is 20.0 V (3 significant figures).
   • Induced current produces flux opposing the decrease, reinforcing the original flux direction.

   Examiner tip: Distinguish flux from flux linkage to avoid multiplying by N twice.

5. 5.

   Parallel rails are 0.400 m apart. A conducting rod spans them perpendicularly and slides along them at 3.00 m s^-1, with the rod perpendicular to its motion. A uniform 0.500 T field is normal to the rail plane. The closed circuit has total resistance 2.00 ohm. Find emf, current, magnetic drag and external power for constant speed. Neglect friction and self-inductance.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. As swept area increases at Lv, flux changes at BLv and drives current around the closed circuit.
   2. Lenz's law gives drag rather than propulsion; an external agent must replace the energy converted to resistive heating.

   Marking points

   • emf = BLv = 0.600 V; I = emf/R = 0.300 A.
   • Drag BIL = 0.0600 N opposite the motion.
   • External power Fv = 0.180 W, equal to I^2R.

   Examiner tip: Constant speed requires an external force balancing the magnetic drag.

6. 6.

   A 100-turn coil of area 0.0100 m^2 rotates uniformly at 50.0 Hz in a uniform 0.200 T field about a fixed axis in the coil plane and perpendicular to the field. Its normal aligns with the field at t = 0. Find peak emf, rms emf and the first positive time of peak magnitude. Neglect losses; give 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The derivative of a cosine flux gives a sine emf; maximum flux at zero means zero emf then.
   2. A sine reaches its first magnitude peak after a quarter period, and its rms is peak divided by sqrt(2).

   Marking points

   • Flux linkage = NBA cos(omega t), so peak emf = NBA omega with omega = 2pi f.
   • Peak emf = 62.8 V; rms emf = 44.4 V.
   • First peak magnitude at T/4 = 0.00500 s.

   Examiner tip: The coil frequency is f, but the peak formula requires angular frequency 2pi f.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.