(a) Ф = BA, read B from the graph at 3 seconds and use the given area (0.2)(0.3) = 0.06 Wb
(b) Induced emf = ε = NΔФ/t … = 1*(BAf – BAi) / t = A(Bf – Bi) / t = (0.3)(0.2 – (–0.2)) / 2 = 0.06 V
(c) First, the directions are found based on Lenz Law. From 0–1 the downward flux is decreasing so current flows CW to add back the downward field. Then we hit zero flux at 1 second. Moving 1–2 seconds we want to maintain the zero flux and we have an increasing upwards flux so the current still flows CW to add downward field to cancel the gaining flux. At 2 seconds the flux becomes constant so current does not flow up until 4 seconds. From 4–6 seconds we are loosing upwards flux so current flows CCW to add back that upwards field.
To determine the current magnitudes. Use V=IR.
From 0–2 sec. 0.06 = I (0.2) I = 0.3 A
From 2–4, I=0.
From 4–6, first determine new emf. Since the slope is half as much as 0–2 sec, the emf should be half as much as well. Then V=IR … 0.03 = I (0.2) I = 0.15 A