Electromagnetic induction for IB Physics HL

Reading cue. The pole facing the ring is N, not the S drawn at the top; as the magnet falls the downward flux through the ring is rising, and that rise fixes the induced current.

1. aState State which pole of the magnet faces the ring, and state the direction of the magnetic field along the dashed line just below the magnet.

   recall2 marks

   Check answer · 2 marks

   1. the north pole faces the ring, the south pole being at the top, away from it
   2. the field just below the magnet points downward, away from the north pole, along the dashed line

2. bDetermine Determine the direction of the induced current in the copper ring as the magnet approaches, as seen by an observer looking down from above the magnet.

   routine3 marks

   Check answer · 3 marks

   1. the flux through the ring points downward and is increasing as the magnet falls
   2. by Lenz's law the induced current opposes the increase, so it must produce upward flux inside the ring
   3. the current therefore flows anticlockwise as seen from above, making the upper face of the ring behave as a north pole

3. cExplain As the magnet falls past the ring the induced current reverses direction, yet the force the ring exerts on the magnet stays upward throughout. Explain why the reversal does not reverse the force, and identify the one position of the magnet at which the ring exerts no force on it at all.

   demanding4 marks

   Show a hint

   Ask where the flux through the ring is greatest, not where the field is strongest.

   Check answer · 4 marks

   1. above the ring the downward flux is increasing, so the current opposes the increase and the upper face of the ring acts as a north pole, repelling the approaching magnet
   2. below the ring the downward flux is decreasing, so the current reverses and the lower face acts as a south pole, attracting the receding magnet
   3. the induced effect always opposes the change producing it, so the force acts against the motion in both phases, that is upward, and the acceleration is less than g
   4. the force is zero when the centre of the magnet is level with the plane of the ring: the flux is a maximum there, so its rate of change is momentarily zero and no current flows

4. dDiscuss Discuss what would change if the copper ring were cut so that a narrow gap ran through it, and separately what would change if the ring were replaced by one of identical dimensions made from a metal of higher resistivity.

   top of the paper4 marks

   Check answer · 4 marks

   1. with a gap the circuit is broken, so although an emf is still induced around the ring, no current can flow
   2. with no current there is no opposing magnetic field and no retarding force, so the magnet falls with acceleration g throughout
   3. with a ring of higher resistivity the same emf drives a smaller current, since I = emf / R, so the retarding force is smaller and the magnet is slowed less
   4. the retarding force is what converts the magnet's gravitational potential energy into resistive heating in the ring, so less current means less heating and a faster arrival

Reading cue. Two slip rings, not a split ring, so the output alternates; and it is the plane of the coil, not its normal, that the drawing tilts relative to the four field arrows.

1. aState State the frequency and the period of the potential difference appearing at the terminals marked as the output to the external circuit, and state whether that potential difference reverses in sign.

   recall3 marks

   Check answer · 3 marks

   1. Frequency = 50 Hz, taken from the 50 revolutions per second marked at the axis
   2. Period = 1/50 = 0.020 s (20 ms)
   3. Potential difference does reverse in sign each half revolution, because the coil is taken off through two slip rings rather than a split ring

2. bSketch Sketch a graph of the output potential difference against time for two complete revolutions of the coil, taking t = 0 at the instant when the plane of the coil contains the direction of the field arrows. Mark the period on the time axis and mark the two peaks as equal in size and opposite in sign.

   routine4 marks

   Check answer · 4 marks

   1. Sinusoidal curve, symmetrical about the time axis and taking both positive and negative values
   2. Curve at a maximum at t = 0, since the flux linkage is zero and changing fastest at that instant
   3. Two complete cycles drawn, each of period 20 ms, filling 40 ms of the time axis
   4. Positive and negative peaks marked equal in size, as the take-off is through two slip rings

3. cDetermine At the instant drawn, the plane of the coil makes an angle of 30° with the direction of the field arrows. Determine the e.m.f. induced at that instant.

   demanding4 marks

   Check answer · 4 marks

   1. ω = 2π × 50 = 314 rad s⁻¹
   2. Peak e.m.f. = NBAω = 250 × 0.085 × 3.2 × 10⁻³ × 314 = 21.4 V
   3. Angle between the normal to the coil and the field is 90° − 30° = 60°, so e.m.f. = 21.4 sin 60°
   4. e.m.f. = 18.5 V (19 V to two significant figures)

4. dDiscuss The arrangement is now altered so that the dashed axis of rotation lies along the field arrows instead of across them, the coil still turning at 50 revolutions per second about that axis. Discuss the output now obtained.

   top of the paper4 marks

   Check answer · 4 marks

   1. The plane of the coil contains the axis of rotation, so the normal to the coil stays perpendicular to the field throughout the turn
   2. The flux linkage is therefore zero at every instant of the rotation
   3. Since the flux linkage does not change, its rate of change is zero and no e.m.f. is induced
   4. No output is obtained however fast the coil is turned, showing that it is the rate of change of flux linkage, not the motion of the coil in the field, that generates the e.m.f.