Generators and Induction Problems: Quick Revision

A brief note on Generators and Induction Problems. Read it before a test. Try each question at the end before you open its answer.

The AC generator

• Emf of a turning coil: $e = NBA\omega\sin\omega t$. The peak is $e_0 = NBA\omega$, with $\omega = 2\pi f$.
• Zero emf: the plane of the coil is at right angles to the field. The flux is greatest and not changing.
• Greatest emf: the plane lies along the field. The flux is zero and changing fastest.
• Direction: the emf reverses twice in each turn.
• Slip rings give AC. A split-ring commutator gives one direction.
• Energy: the driver's mechanical power $\tau\omega$ equals the electrical power $ei$.
• Steady value: $e_{rms} = \dfrac{e_0}{\sqrt{2}}$. The average power in a resistor is $\dfrac{e_0^2}{2R}$.

Rods and frames

• Four steps: $e = Blv$, then $I = \dfrac{e}{R}$, then $F = BIl$ (against the motion), then Newton's law or energy.
• A rod with a resistor: $F = \dfrac{B^2l^2v}{R}$ and the power is $\dfrac{B^2l^2v^2}{R}$.
• A falling rod: $v_t = \dfrac{mgR}{B^2l^2}$ and $\tau = \dfrac{mR}{B^2l^2}$.
• A frame crossing a field: emf and force only while it enters and while it leaves. Heat in all: $\dfrac{2B^2a^3v}{R}$.
• Charge that flows: $q = \dfrac{\Delta\Phi}{R}$, whatever the speed.
• With a capacitor: an extra mass $CB^2l^2$, and a constant acceleration $a = \dfrac{mg}{m + CB^2l^2}$.
• A rod on a battery: back emf $Blv$, and $I = \dfrac{\mathcal{E} - Blv}{R}$. It stops speeding up at $v = \dfrac{\mathcal{E}}{Bl}$.

The LR circuit

• Time constant: $\tau = \dfrac{L}{R}$, in seconds.
• Switching on: $I = \dfrac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right)$. At $t = 0$ the coil acts like a break. After a long time it acts like a wire.
• Dying away: $I = I_0e^{-t/\tau}$. After $\tau$ the current is $37\ \%$ of the start.
• Time to a fraction $f$: $t = -\tau\ln(1 - f)$.
• Breaking a current at once: a very large emf, $L\dfrac{\Delta I}{\Delta t}$, and a spark.
• Energy given back: $\tfrac{1}{2}LI_0^2$, as heat in the resistor.

A worked example

A square frame of side $5.0$ cm and resistance $0.020\ \Omega$ enters a field of $0.30$ T at $0.40$ m/s. Find the current, and the force on the frame while it enters.

1. The emf. $e = Bav = 0.30\ \text{T} \times 0.050\ \text{m} \times 0.40\ \text{m/s} = 0.0060\ \text{V}$.
2. The current. $I = \dfrac{e}{R} = \dfrac{0.0060\ \text{V}}{0.020\ \Omega} = 0.30\ \text{A}$.
3. The force. $F = BIa = 0.30\ \text{T} \times 0.30\ \text{A} \times 0.050\ \text{m} = 0.0045\ \text{N}$, against the motion.

Practise with new numbers

Check yourself

1. A generator turns twice as fast. By what factor does the average power in a fixed resistor change?

2. A coil has $L = 0.90$ H and a total resistance of $6.0\ \Omega$. What is its time constant?

3. A rod falls on rails with a resistor. Why does it stop speeding up?