Electromagnetism

Reading cue. The arrows leave the north pole and enter the south, so the field runs left to right; the force is perpendicular to both field and wire, so it cannot lie in the plane of the page.

1. aDescribe Describe what happens to the force on the wire in Fig. 5.1 if the current I is reversed so that it flows down the page instead of up.

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   Check answer · 2 marks

   1. the force acts in the opposite direction
   2. its size is unchanged, because neither the current nor the strength of the field has been altered

2. bDescribe The two magnets in Fig. 5.1 are now exchanged, so that a south pole faces the gap from the left and a north pole from the right, and at the same time the current is reversed. Describe the effect of these two changes together on the force on the wire.

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   Check answer · 3 marks

   1. reversing the field alone would reverse the force, and reversing the current alone would reverse it as well
   2. with both reversed the two changes cancel each other
   3. the force therefore acts in the same direction as before and has the same size

3. cExplain The wire in Fig. 5.1 is turned slowly in the plane of the page, away from the vertical position drawn, until it finally lies horizontally along the field arrows. Explain how the size of the force on the wire changes as it is turned, and state the position in which the force is largest.

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   Check answer · 4 marks

   1. the force is largest in the position drawn in Fig. 5.1, with the wire at right angles to the field arrows
   2. as the wire is turned away from that position the force becomes steadily smaller
   3. when the wire lies along the field arrows the force is zero, because no part of the current then crosses the field
   4. throughout the turning the force stays perpendicular to the page, so only its size changes

4. dSuggest Suggest three separate changes to the arrangement in Fig. 5.1, each of which would increase the size of the force on the wire, and explain why only the length of wire lying between the pole faces affects that force.

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   Check answer · 4 marks

   1. increase the current in the wire
   2. use stronger magnets, or bring the pole faces closer together, so that the field across the gap is stronger
   3. increase the length of wire lying in the field, for example by using wider pole faces or by replacing the single wire with several wires side by side carrying the current the same way
   4. outside the gap the field is very weak, so the parts of the wire beyond the pole faces experience almost no force and do not contribute

Reading cue. Take the turns from the labels, not from the loops drawn: five on the left and seven on the right, but the ratio that decides the output is 200 to 5000.

1. aIdentify Identify which coil in Fig. 6.1 has the greater number of turns, and state whether the transformer is a step-up or a step-down transformer.

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   Check answer · 2 marks

   1. the coil wound on the right-hand limb, the secondary, with 5000 turns
   2. it is a step-up transformer, because the secondary has more turns than the primary

2. bExplain The 12 V a.c. supply in Fig. 6.1 is replaced by a 12 V d.c. supply. Explain what a voltmeter connected across the output terminals would read.

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   Check answer · 3 marks

   1. the voltmeter reads zero, apart from a momentary reading as the supply is switched on or off
   2. a steady direct current produces a steady magnetic field in the iron core
   3. there is then no change of magnetic field through the secondary coil, so no e.m.f. is induced in it

3. cDetermine The output terminals in Fig. 6.1 are connected to a lamp, and the current in the primary coil is 0.50 A. Assuming the transformer is 100% efficient, determine the current in the secondary coil and the power delivered to the lamp.

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   Check answer · 3 marks

   1. input power = 12 × 0.50 = 6.0 W, and at 100% efficiency the lamp receives 6.0 W
   2. secondary current = primary current × 200/5000 = 0.50 × 0.040
   3. current in the secondary = 0.020 A (20 mA)

4. dExplain The core in Fig. 6.1 is drawn as a solid rectangle of iron. In a real transformer it is built from thin sheets separated by insulation. Explain how this changes what happens in the core, and describe one further reason why a real transformer is not 100% efficient.

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   Check answer · 4 marks

   1. the changing magnetic field induces currents in the iron of the core itself
   2. in a solid core these currents circulate freely and heat the core, so energy from the supply is wasted
   3. insulated sheets break up the paths available to these currents, so they are much smaller and less energy is wasted
   4. one further loss: the copper coils have resistance, so the current in them heats the windings (accept: not all the field from the primary passes through the secondary, or energy is wasted repeatedly magnetising the core)

Reading cue. Read the pole from the end of the magnet nearest the coil, not the labelled far end, and note the arrow sits on the magnet: the magnet moves and the coil stays still.

1. aIdentify Identify the feature of the meter drawn in Fig. 8.1 that lets the direction of the induced current be found, and state what its needle reads before the magnet is moved.

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   Check answer · 3 marks

   1. it is a centre-zero ammeter / zero is at the middle of the scale (1)
   2. the needle can swing either side of zero, so the direction of the current is shown (1)
   3. reads zero before the magnet moves (1)

2. bState For the motion drawn by the arrow in Fig. 8.1, state which magnetic pole is produced at the end of the coil nearest the magnet, and state the effect this has on the magnet as it moves in.

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   Check answer · 2 marks

   1. the near end of the coil becomes a north pole (1)
   2. it repels the approaching north pole of the magnet / opposes the magnet's motion (1)

3. cExplain The magnet in Fig. 8.1 is now held still and the coil is moved to the left towards it, at the same speed. Explain what the ammeter shows.

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   Check answer · 4 marks

   1. the needle deflects in the same direction as before (1)
   2. by the same amount (1)
   3. only the relative movement of magnet and coil matters (1)
   4. the field through the coil changes at the same rate, so the same e.m.f. is induced (1)

4. dExplain The two wires running down from the coil in Fig. 8.1 are disconnected from the ammeter and the magnet is pushed in again at the same speed. Explain why less force is now needed to push the magnet in, and state the source of the energy that was previously measured as a current.

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   Check answer · 4 marks

   1. with the circuit complete an induced current flows in the coil (1)
   2. this current makes the coil into a magnet whose near pole repels the incoming north pole, so a force must be overcome (1)
   3. with the wires disconnected the circuit is broken, so no current flows and there is no opposing force (an e.m.f. is still induced) (1)
   4. the electrical energy came from the work done by the person pushing the magnet / from the magnet's kinetic energy (1)

Reading cue. F is marked on both sides of the lens, and the object stands 10 cm away against f = 15 cm: it is inside the focal point, and that single fact decides the whole answer.

1. aSketch Sketch on a copy of the figure the ray that leaves the tip of the object travelling parallel to the principal axis, and indicate its path after it has passed through the lens.

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   Check answer · 2 marks

   1. a straight ray drawn from the tip of the object arrow, parallel to the dashed axis, as far as the lens
   2. after the lens the ray is bent towards the axis and drawn through the focal point F marked on the far side

2. bDetermine Determine the image distance for the object position marked on the figure.

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   Check answer · 3 marks

   1. 1/v = 1/f − 1/u = 1/15 − 1/10
   2. 1/v = (2 − 3)/30 = −1/30
   3. v = −30 cm, so the image lies 30 cm from the lens on the same side as the object

3. cCalculate The object arrow drawn on the figure represents an object 4.0 mm tall. Calculate the height of the image and determine the distance between the object and its image.

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   Check answer · 4 marks

   1. magnification m = −v/u = −(−30)/10 = +3.0
   2. the positive sign shows the image is upright, the same way up as the drawn object arrow
   3. image height = 3.0 × 4.0 = 12 mm
   4. both lie on the same side of the lens, so the separation is 30 − 10 = 20 cm

4. dDetermine Determine how far the object must be moved along the axis for the image to lie twice as far from the lens as it does now, and explain what happens as the object is moved all the way out to the focal point.

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   Check answer · 4 marks

   1. for the image at v = −60 cm, 1/u = 1/f − 1/v = 1/15 + 1/60
   2. 1/u = 5/60, so u = 12 cm
   3. the object must be moved 12 − 10 = 2.0 cm further from the lens, that is towards the near focal point
   4. as u approaches 15 cm, 1/v approaches zero and the image distance grows without limit: the rays leave the lens parallel and no image is formed at all

Reading cue. Both dimension lines run to the centre of the lens, not to its mount or its glass face, so a misjudged centre raises u by exactly what it takes off v.

1. aDetermine With a sharp image on the screen, the object stands at the 12.0 cm mark on the rail, the lens at the 42.0 cm mark and the screen at the 102.0 cm mark. Determine u and v.

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   Check answer · 2 marks

   1. u is the object-to-lens distance: 42.0 − 12.0 = 30.0 cm
   2. v is the lens-to-screen distance: 102.0 − 42.0 = 60.0 cm

2. bCalculate Calculate the magnification of the image for that setting, and determine whether the image caught on the screen is upright or inverted.

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   Check answer · 3 marks

   1. magnification m = −v/u = −60.0/30.0
   2. m = −2.0, so the image is twice the height of the illuminated object
   3. an image that can be caught on a screen is real, and the negative sign shows it is inverted

3. cDetermine Leaving the object and the screen exactly where they are, determine the second position of the lens on the rail that also gives a sharp image, and state the magnification there.

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   Check answer · 4 marks

   1. from part (a), 1/f = 1/30.0 + 1/60.0 = 1/20.0, so f = 20.0 cm
   2. object and screen are fixed 90.0 cm apart, so u + v = 90.0 while uv = f(u + v) = 20.0 × 90.0 = 1800 cm²
   3. u and v are then the roots of t² − 90t + 1800 = 0, namely 60.0 cm and 30.0 cm, so the lens also focuses when set at the 72.0 cm mark
   4. the magnification there is −30.0/60.0 = −0.50, the reciprocal in size of the value in part (b)

4. dExplain The student now moves the screen so that object and screen are only 70.0 cm apart. Explain why no position of this lens on the rail will give a sharp image, and determine the smallest separation that does allow one.

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   Show a hint

   The two lens positions in part (c) came out of a quadratic. Ask what happens to the roots of that quadratic as the object and the screen are brought closer together.

   Check answer · 4 marks

   1. with a separation D the lens positions satisfy t² − Dt + fD = 0, whose roots are real only if D² ≥ 4fD, that is D ≥ 4f
   2. for this lens 4f = 80.0 cm, so 70.0 cm is too small a separation
   3. at D = 70.0 cm the discriminant is 70² − 4(20)(70) = 4900 − 5600 = −700, which is negative, so there is no real lens position at all
   4. the smallest workable separation is 80.0 cm, where the two positions merge into one at the midpoint with u = v = 40.0 cm and magnification 1 in size

Reading cue. Crosses cover the page inside and outside the loop, but only those inside it carry flux; the field is into the page and growing, so the induced current opposes the growth, not the field.

1. aDetermine The loop has radius a and the field magnitude is B. Determine the magnetic flux through the loop, and state why the crosses drawn outside the loop make no difference to it.

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   Check answer · 3 marks

   1. Φ = Bπa²
   2. the field is uniform and perpendicular to the plane of the loop, so no angle factor is needed
   3. flux counts only the field crossing the area enclosed by the loop

2. bCalculate The loop has radius 6.0 cm and resistance 0.25 Ω, and the field increases steadily from 0.20 T to 0.50 T in 1.5 s. Calculate the induced e.m.f. and the current in the loop.

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   Check answer · 4 marks

   1. A = π(0.060)² = 1.13 × 10⁻² m²
   2. dB/dt = 0.30/1.5 = 0.20 T s⁻¹
   3. e.m.f. = A dB/dt = 2.3 × 10⁻³ V
   4. I = e.m.f./R = 9.0 × 10⁻³ A

3. cDetermine Determine the direction of the magnetic force on the wire at the top of the loop, and hence determine whether the wire of the loop is squeezed inwards or stretched outwards while the field is increasing.

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   Check answer · 5 marks

   1. the induced current is counterclockwise, opposing the increasing into-page flux
   2. at the top of the loop that current points to the left
   3. F = IL × B on that element points towards the centre of the loop
   4. the same argument holds all round, so the loop is squeezed inwards and the wire is in compression
   5. consistent with Lenz's law, since shrinking the area would reduce the flux that is increasing

4. dExplain The loop is cut so that a narrow gap is left, and an ideal voltmeter is connected across the gap. Explain which of the flux, the e.m.f., the current, the dissipated power and the force on the wire change, and state the voltmeter reading.

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   Check answer · 5 marks

   1. the flux and the e.m.f. are unchanged, since both depend on the field and the enclosed area and not on the wire
   2. the current falls to zero, because the conducting path is broken
   3. the dissipated power falls to zero
   4. the magnetic force on the wire disappears, since it required a current
   5. the voltmeter reads the full induced e.m.f., 2.3 × 10⁻³ V

Reading cue. L is tick-marked between the rails, so it is the length of rod in the circuit, not a distance travelled; the loop's area is bounded by R at one end and the moving rod at the other.

1. aIndicate Indicate the direction, clockwise or counterclockwise as drawn, of the induced current around the loop while the rod moves to the right.

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   Check answer · 3 marks

   1. the enclosed area is growing, so the outward flux through the loop is increasing
   2. the induced current opposes this by producing flux into the page inside the loop
   3. the current is therefore clockwise as drawn, passing down through the rod

2. bCalculate Take B = 0.45 T, L = 0.35 m, R = 0.80 Ω, m = 0.15 kg and v₀ = 4.0 m s⁻¹. Calculate the magnetic force on the rod and its deceleration at t = 0.

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   Check answer · 4 marks

   1. e.m.f. = BLv₀ = 0.45 × 0.35 × 4.0 = 0.63 V
   2. I = e.m.f./R = 0.79 A
   3. F = BIL = 0.124 N
   4. a = F/m = 0.83 m s⁻², directed opposite to the motion

3. cDetermine Determine the speed of the rod 1.0 s after release, the distance it has travelled in that time, and the energy dissipated in R during that second.

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   Check answer · 4 marks

   1. time constant τ = mR/B²L² = (0.15 × 0.80)/(0.1575)² = 4.8 s
   2. v = v₀e^(−t/τ) = 4.0 × e^(−0.207) = 3.3 m s⁻¹
   3. distance = v₀τ(1 − e^(−t/τ)) = 4.0 × 4.84 × 0.187 = 3.6 m
   4. energy = ½m(v₀² − v²) = ½ × 0.15 × (16 − 10.6) = 0.41 J, all of it dissipated in R

4. dDerive Starting from rest, the rod is now pushed to the right by a constant applied force of 0.50 N. Derive an expression for its terminal speed, calculate its value, and show that at that speed every joule supplied by the applied force is dissipated in R.

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   Check answer · 5 marks

   1. equation of motion m dv/dt = F − B²L²v/R
   2. at terminal speed dv/dt = 0, so v_T = FR/B²L²
   3. v_T = (0.50 × 0.80)/(0.1575)² = 16 m s⁻¹
   4. mechanical power supplied = Fv_T = 8.1 W
   5. electrical power dissipated = (BLv_T)²/R = 8.1 W, equal because the kinetic energy is no longer changing

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.

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   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.

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   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.

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   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.

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   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.

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   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.

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   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.

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   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.

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   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.