Notes
Magnets and fields
- Like poles repel, unlike attract; poles always come in N-S pairs. Field lines run from north to south outside a magnet.
- A current makes a magnetic field. An electromagnet (coil plus iron core) is stronger with more current, more turns or an iron core.
- Force on a wire perpendicular to a field: $F = BIL$. Motors change electrical energy into mechanical energy.
Induction
- A changing magnetic field induces a voltage. Generators change mechanical energy into electrical energy.
- Transformers: $\displaystyle \frac{V_s}{V_p} = \frac{N_s}{N_p}$; they only work with AC.
The nucleus and radioactivity
- $Z$ = protons, $A$ = protons + neutrons. Isotopes differ in neutrons.
- Alpha (helium nucleus, stopped by paper), beta (electron, stopped by aluminum), gamma (EM wave, stopped by thick lead).
- After $n$ half-lives, $\displaystyle \left(\frac{1}{2}\right)^n$ of the sample remains.
Nuclear energy
- $E = mc^2$. Fission splits heavy nuclei (power plants); fusion joins light nuclei (the Sun).
Topics in this unit
Each topic has its own short notes and practice.
Practice
Like poles repel.
$F = BIL = 0.25 \times 5 \times 0.4 = 0.5 \text{ N}$.
$\displaystyle N_s = 50 \times \frac{120}{12} = 500$.
Induction requires a changing magnetic field.
11,460 years is 2 half-lives: $\displaystyle \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
$Z = 27$ protons; neutrons $= 60 - 27 = 33$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A small wind turbine spins a coil of wire inside a magnetic field. Its AC output of 240 V goes into a transformer with 1,200 turns on the primary coil and 60 turns on the secondary coil, which charges equipment that runs at 12 V.
(a) Explain how the turbine produces electricity. Name the energy change.
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The spinning coil keeps changing the magnetic field through it, which induces a voltage and current (electromagnetic induction). Kinetic (mechanical) energy of the wind becomes electrical energy.
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(b) Calculate the transformer's output voltage. Is it step-up or step-down?
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$\displaystyle V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{60}{1200} = 12 \text{ V}$. It is a step-down transformer because the secondary has fewer turns.
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(c) The equipment uses 2.0 A at 12 V. Assuming an ideal (100% efficient) transformer, find the power delivered and the current in the primary coil.
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$P = IV = 2.0 \times 12 = 24 \text{ W}$. For an ideal transformer, power in $=$ power out, so $\displaystyle I_p = \frac{24}{240} = 0.10 \text{ A}$.
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