1. A square coil consists of 150 turns of wire. Each turn measures 12cm by 15cm. The coil is placed in a magnetic filed of 0.25T. Do now: Mane, 16 Whiringa-ā-rangi 2015 1) Calculate the total flux threading the coil 2) The field dies to zero in a time of 12ms. Calculate the voltage induced across the solenoid.

2. Today’s Learning: Mane, 16 Whiringa-ā-rangi 2015 1) What is inductance? 2) What energy is stored in an inductor? Homework: Capacitor questions due in (yesterday)

3. The flux is proportional to the current: Current in an inductor produces magnetic flux: Unit of inductance: Henry (H) The constant of proportionality is the inductance, L Key concept: as the current changes, the flux changes, which induces an opposing voltage (back EMF) 1H is large! Typically: nH, μH, mH

4. Derive: An inductor only affects circuits when the current is changing and will always oppose what is happening The size of the induced back EMF depends on: 1) the inductance of the inductor and 2)the rate of change of current An inductor causes a delay in a lamp reaching its full brightness, because energy is stored in the induced magnetic field of the inductor, rather than causing the lamp to glow Energy stored in an inductor:

5. Watch clips: 1) Inductor cooktop experiments - 1min 30sec 2) LabTV – Inductors – 4min 15sec 3) Inductors - good but a bit technical - 3min 50sec Keep in mind: 1) A capacitor stores energy in an electric field, an inductor stores energy in a magnetic field 2) An inductor only affects a circuit while the current is changing i.e. in an AC circuit or a DC circuit being switched on or off – if the current is steady, then the inductor acts just like a resistor

6. The coil has an inductance of 0.20H and the supply voltage is 3.0V. When the switch opens, the current falls from 5.0A to zero in 0.010s. a) What is the induced voltage? b) What is the resistance of the coil? c) What is the energy stored in the coil?

7. Solutions to worksheet: 1(a)Ø = B x A = 2.4 x 10 -2 x 0.033 = 7.9 x 10 -4 Wb (b)V = ∆Ø/∆t = 7.9 x 10 -4 /3.5 = 2.3 x 10 -4 V 2(a)A = anticlockwise, B = no current, C = clockwise (b)Ø = B x A = 0.85 x (0.12 x 0.12) = 1.2 x 10 -2 Wb (c)v = d/t so t = d/v = 12/3 = 4 s (d)V = ∆Ø/∆t = 1.2 x 10 -2 /4 = 3.0 x 10 -3 V 3 (a)The light bulb with the inductor lights up more slowly because as the current changes through the inductor (zero to maximum), a back Emf is induced in the inductor which opposes the current producing it. This causes the forward current to build up more slowly to its maximum steady value.