The flux through each turn of a 50-turn coil changes from 0.3 Wb to 0.8 Wb in 0.4 s. Find the average induced emf.

Solution & explanation

1. Given: N = 50; Φ₁ = 0.3 Wb; Φ₂ = 0.8 Wb; Δt = 0.4 s.

2. Subtract initial flux from final flux.

   $${\Delta }_{\Phi }=0.8-0.3=0.5$$

3. Faraday’s law includes Lenz’s minus sign.

   $$\epsilon =\frac{-{\Delta }_{\Phi }\times N}{{\Delta }_t}$$

4. Insert the signed flux change.

   $$\epsilon =-\frac{25.0}{0.4}$$

5. Multiply turns by flux change.

   $$\epsilon =-\frac{25}{0.4}=-62.5$$

6. State the result with units: Magnitude 62.5 V; −62.5 V relative to the chosen positive flux orientation.

Average emf magnitude = 62.5 V