Part of MAG-02 — Electromagnetic Induction & Alternating Current

Worked Problems with Units

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Problem 1: Motional EMF with Force

Given: Rod length l = 50 cm = 0.50 m, velocity v = 4 m/s, field B = 0.5 T, resistance R = 2 Ω\Omega.

Step 1 — Motional EMF: ε=Bvl=0.5T×4m/s×0.50m=1.0V\varepsilon = Bvl = 0.5\,\text{T} \times 4\,\text{m/s} \times 0.50\,\text{m} = 1.0\,\text{V}

Step 2 — Induced current: I=εR=1.0V2Ω=0.5AI = \frac{\varepsilon}{R} = \frac{1.0\,\text{V}}{2\,\Omega} = 0.5\,\text{A}

Step 3 — Force on rod (Lenz's law: opposing motion): F=BIl=0.5T×0.5A×0.50m=0.125NF = BIl = 0.5\,\text{T} \times 0.5\,\text{A} \times 0.50\,\text{m} = 0.125\,\text{N}

Step 4 — Power check (energy conservation): Pexternal=Fv=0.125×4=0.5WP_\text{external} = Fv = 0.125 \times 4 = 0.5\,\text{W} Pdissipated=I2R=(0.5)2×2=0.5WP_\text{dissipated} = I^2 R = (0.5)^2 \times 2 = 0.5\,\text{W} \checkmark

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