Title: InkPhysics / Magnetic fields
On every note of this deck, unless a note says otherwise:
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Front: CHECK YOURSELF
An electron moves at 2.0 × 10^7 m s^−1 at right angles to a field of 0.50 mT. Find the force on it and the radius of its circular path.
Show a hint
BQv for the force; then let it be the centripetal force.
Back: F = BQv = 5.0 × 10^−4 × 1.60 × 10^−19 × 2.0 × 10^7 = 1.6 × 10^−15 N.
r = mv/BQ = (9.11 × 10^−31 × 2.0 × 10^7) / (5.0 × 10^−4 × 1.60 × 10^−19) = 0.23 m.
The speed stays 2.0 × 10^7 m s^−1 the whole way round: the field steered the electron without giving it a single joule.

Front: DEFINITION
Define: Root-mean-square value.
Back: The value of the steady direct current that would dissipate the same average power in a resistor: I_rms = I₀/√2.

Front: EQUATION — NOT ON THE DATA SHEET
Radius of a charge's circular path in a field:
r = ?
Back: r = mvBQ

Front: DEFINITION
Define: Flux linkage.
Back: The flux through a coil multiplied by its number of turns: NΦ.

Front: CHECK YOURSELF
A station sends 10 MW down cables of total resistance 5.0 Ω. Find the power lost when transmitting at 25 kV, and at 400 kV, as a percentage of the power sent each time.
Show a hint
Current first from P = IV; then the loss is that current squared times R.
Back: At 25 kV: I = P/V = 10^7/(2.5 × 10^4) = 400 A, so loss = I^2R = 400^2 × 5.0 = 8.0 × 10^5 W: 0.80 MW, 8.0% gone.
At 400 kV: I = 25 A, so loss = 25^2 × 5.0 = 3.1 × 10^3 W: 3.1 kW, 0.031%.
Sixteen times the voltage, 256 times less loss: the square at work, and the reason pylons carry hundreds of kilovolts.

Front: DEFINITION
Define: Faraday's law.
Back: The induced emf is equal to the rate of change of flux linkage.

Front: DEFINITION
Define: Lenz's law.
Back: The induced current flows in the direction that opposes the change of flux producing it.

Front: CHECK YOURSELF
A 500-turn coil of area 2.0 × 10^−3 m² sits in a 0.15 T field with its normal at 60° to the field. Find the flux linkage, and state what it becomes face-on.
Show a hint
BANcosθ, with θ to the normal; then set θ to zero.
Back: NΦ = BANcos θ = 0.15 × 2.0 × 10^−3 × 500 × cos 60° = 0.075 Wb turns.
Face-on, cosθ = 1, so the linkage doubles to 0.15 Wb turns.
Edge-on it would fall to zero: the two limits that catch any misread angle.

Front: CHECK YOURSELF
An oscilloscope shows a sine wave of amplitude 3.0 divisions with one cycle spanning 4.0 divisions. The y-gain is 5.0 V per division and the timebase 2.0 ms per division. Find the peak voltage, the rms voltage, and the frequency.
Show a hint
Squares to volts and seconds first; the physics is two multiplications and a reciprocal.
Back: Peak: V_0 = 3.0 × 5.0 = 15 V.
V_rms = V_0/√2: 15 / 1.414 = 10.6 V.
Period T = 4.0 × 2.0 ms = 8.0 ms, so f = 1/T = 125 Hz.

Front: DEFINITION
Define: Magnetic flux density.
Back: The force per unit current per unit length on a wire carrying current at right angles to the field: B = F/IL, in tesla.

Front: CHECK YOURSELF
A wire carries 3.0 A at right angles across a field of 0.20 T, with 5.0 cm of the wire in the field. Find the force, and the change in reading of a balance supporting the magnet, in grams.
Show a hint
F = BIl in base units, then grams via g.
Back: F = BIl = 0.20 × 3.0 × 0.050 = 0.030 N.
The magnet feels the equal and opposite partner force, so the reading changes by F/g = 0.030 / 9.81 = 3.1 × 10^−3 kg, about 3.1 g.
Whether the reading rises or falls depends on the current's direction; reversing it flips the change, which is itself a check the effect is real.

Front: CHECK YOURSELF
A 200-turn coil of area 1.5 × 10^−3 m² rotates at 50 revolutions per second in a 0.20 T field. Find the peak emf, and state the emf at the instant the coil is face-on to the field.
Show a hint
Convert revolutions to radians first; then the peak is everything in front of the sine.
Back: ω = 2π × 50 = 314 rad s^−1.
Peak ε = BANω = 0.20 × 1.5 × 10^−3 × 200 × 314 = 19 V.
Face-on the linkage is maximal but momentarily unchanging, so the emf is zero: the quarter-turn offset in action.

Front: DEFINITION
Define: Magnetic flux.
Back: The product of flux density and the area it passes through at right angles: Φ = BA, in webers.