A 25-turn coil has a constant induced emf of 2.5 V for 0.20 s. Find the magnitude of change in magnetic flux through each turn.

Solution & explanation

1. Given N = 25, emf magnitude 2.5 V and time 0.20 s.

2. Use the magnitude of Faraday’s law.

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

3. Multiply by Δt and divide by N.

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

4. Substitute.

   $${\Delta }_{\Phi m}=\frac{0.5}{25}$$

5. Multiply the numerator, then divide.

   $${\Delta }_{\Phi m}=\frac{0.5}{25}=0.02$$

6. This is flux per turn, not NΦ flux linkage; sign needs an orientation.

abs(ΔΦ) = 0.020 Wb per turn