B = 1T, V = 5I10–2m/', R = 10Ω
(a) When the switch is thrown to the middle rail
E = Bvℓ
= 1 × 5 × 10–2 × 2 × 10–2 = 10–3
Current in the 10 resistor = E/R
= 10-3/10 = 10-4 = 0.1mA
(b) The switch is thrown to the lower rail
E = Bvℓ
= 1 × 5 × 10–2 × 2 × 10–2 = 20 × 10–4
current = 20 x 10-4/10 = 2 x 10-4 = 0.2mA