Grade 11 Science · Problem type

Flux from a changing field

Use this for “A 200-turn coil of area 0.02 m² sits in a field that grows from 0.1 T to 0.5 T in 0.1 s. What emf is induced?”

Diagram

SNpushed inN = 200 turns (20 drawn)Gemf 16 V
emf = NΔΦ/Δt = 200 × 0.008/0.1 = 16 V
Pushing the N pole in, the coil’s current makes its near end an N pole that pushes back (Lenz’s law).
Faster motion or more turns: a bigger emf.
No motion: no current.
Field at the start: 0.1 T
Field at the end: 0.5 T
Area of the coil: 0.02 m²
This is an example. Type your own values to start a new problem.
ΔΦ (Wb)
=
(
B₂ (T)
−
B₁ (T)
)
×
A (m²)
emf (V)
=
N
×
ΔΦ (Wb)
Δt (s)

Formulas

ΔΦ
=
(
B₂
−
B₁
)
×
A
0.008 = (0.5 − 0.1) × 0.02
emf
=
N
×
ΔΦ/Δt
16 = 200 × 0.008/0.1

Assumptions

•
The field is square to the coil’s face, so the flux is Φ = BA.
•
The area stays the same, so the flux changes by ΔΦ = (B₂ − B₁)A.
•
The field grows here; a shrinking one gives a negative ΔΦ and an emf the other way (Lenz’s law).

Step-by-step

WE KNOW
N = 200, B₁ = 0.1 T, B₂ = 0.5 T, A = 0.02 m², Δt = 0.1 s
FIND
change in flux (ΔΦ), induced emf (emf)
Step 1 · Find change in flux (ΔΦ)
Use ΔΦ = (B₂ − B₁) × A
The change in the field times the area it passes through.
ΔΦ = (0.5 − 0.1) × 0.02
ΔΦ = 0.4 × 0.02
ΔΦ = 0.008 Wb
Step 2 · Find induced emf (emf)
Use emf = N × ΔΦ/Δt
Each
turn
gets
ΔΦ/Δt;
N
turns
add
up.
emf = 200 × 0.008/0.1
emf = 1.6/0.1
emf = 16 V
Check
Put the numbers back in. Both sides should match.
0.008 = (0.5 − 0.1) × 0.02 ✓
16 = 200 × 0.008/0.1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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