Electromagnetic induction for AP Physics C: Electricity and Magnetism

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