A rectangular loop of dimensions 0.2 m by 0.3 m and resistance 2 ohms is pulled at constant speed 5 m/s out of a uniform magnetic field B = 1.5 T perpendicular to the loop. What external mechanical power is required?

Answer and explanation

Correct answer: D. 1.125 W

  1. Step 1: The width perpendicular to motion is L = 0.2 m (or 0.3 m). Let L = 0.2 m be the edge cutting field lines.
  2. Step 2: Induced EMF: E = B L v = 1.5 0.2 5 = 1.5 V.
  3. Step 3: Induced current: I = E / R = 1.5 / 2 = 0.75 A.
  4. Step 4: Magnetic retarding force: F = I L B = 0.75 0.2 1.5 = 0.225 N.
  5. Step 5: Mechanical power required: P = F v = 0.225 5 = 1.125 W (or P = E^2 / R = 1.5^2 / 2 = 2.25 / 2 = 1.125 W).
Answer
1.125 W.
Why not the others: 2.25 W computes E^2 without dividing by R; 0.56 W halves the power; 4.50 W computes P = E^2 * 2.

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