Faraday's law and induced emf: Class 12 practice
Induced emf = −N dΦ/dt where Φ = BA cos θ. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Area
- Rates of change
The idea behind faraday's law and induced emf
Induced emf = −N dΦ/dt where Φ = BA cos θ. A rod of length l moving at v perpendicular to B has emf Blv. The minus sign (Lenz's law) means the current opposes the change.
Step-by-step method
- Find the change in flux ΔΦ = Δ(BA cos θ).
- emf = NΔΦ/Δt (magnitude).
- For a moving rod use Blv.
Worked example 1 (Basic)
A 0.4 m rod moves at 8 m/s perpendicular to a 0.8 T field. Find the emf.
- ε = Blv = 0.8 × 0.4 × 8
- = 2.56 V
- Answer: 2.56 V
Worked example 2 (Standard)
A 97-turn coil of area 0.011 m² is in a 0.9 T field, which falls to zero in 0.2 s. Find the average emf.
- ΔΦ per turn = 0.9 × 0.011 = 0.01 Wb
- ε = NΔΦ/Δt = 4.801 V
- Answer: 4.801 V
Worked example 3 (Challenge)
For the same coil (37 turns, 0.038 m², 0.4 T → 0 in 0.3 s) with resistance 9 Ω, find the charge that flows.
- Q = NΔΦ/R
- = 0.062 C
- Answer: 0.062 C
Common mistakes and misconceptions
- Forgetting N turns
- Using A in cm²
Practice questions
- (Basic) A 0.3 m rod moves at 7 m/s perpendicular to a 0.1 T field. Find the emf.
- (Basic) A 0.5 m rod moves at 19 m/s perpendicular to a 0.4 T field. Find the emf.
- (Standard) A 154-turn coil of area 0.046 m² is in a 0.3 T field, which falls to zero in 0.2 s. Find the average emf.
- (Standard) A 23-turn coil of area 0.05 m² is in a 0.2 T field, which falls to zero in 0.6 s. Find the average emf.
- (Challenge) For the same coil (130 turns, 0.035 m², 0.8 T → 0 in 0.6 s) with resistance 10 Ω, find the charge that flows.
- (Challenge) For the same coil (195 turns, 0.038 m², 0.7 T → 0 in 0.1 s) with resistance 6 Ω, find the charge that flows.
Answer key with full working
- Q1: ε = Blv = 0.1 × 0.3 × 7 → = 0.21 V → Answer: 0.21 V
- Q2: ε = Blv = 0.4 × 0.5 × 19 → = 3.8 V → Answer: 3.8 V
- Q3: ΔΦ per turn = 0.3 × 0.046 = 0.014 Wb → ε = NΔΦ/Δt = 10.626 V → Answer: 10.626 V
- Q4: ΔΦ per turn = 0.2 × 0.05 = 0.01 Wb → ε = NΔΦ/Δt = 0.383 V → Answer: 0.383 V
- Q5: Q = NΔΦ/R → = 0.364 C → Answer: 0.364 C
- Q6: Q = NΔΦ/R → = 0.864 C → Answer: 0.864 C
Syllabus note
This skill is listed under "Electromagnetic Induction" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is faraday's law and induced emf in?
"Electromagnetic Induction", Class 12 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.