FARADAY LAW

1. P22- 1
FARADAYS LAW
RAI SAHEB BHANWAR SINGH
COLLEGE (RALA)
NASRULLAGANG
BY – Sonu chouhan
b.sc. (CS) 4TH
sem
roll no. - 151990365

2. P22-
Faraday’s Law
Faraday's law of induction is a
basic law of electromagnetism
predicting how a magnetic field
will interact with an electric circuit
to produce an electromotive force
(EMF)—a phenomenon called
electromagnetic induction.

3. P22-
Electromagnetic Induction

4. P22-
Faraday’s Law of Induction
Bd
dt
ε Φ
= −
A changing magnetic flux
induces an EMF

5. P22-
What is EMF?
dε = ×∫ E s
r r
Ñ
Looks like potential. It’s a
“driving force” for current

6. P22-
Faraday’s Law of Induction
Bd
d
dt
ε Φ
= × = −∫ E s
r r
Ñ
A changing magnetic flux induces
an EMF, a curling E field

7. P22-
Magnetic Flux Thru Wire Loop
cosB B A BA θ⊥Φ = = = ×B A
rr
B
S
d= ×∫∫Φ B A
rr
Analogous to Electric Flux (Gauss’ Law)
(1) Uniform B
(2) Non-Uniform B

8. P22-
Minus Sign? Lenz’s Law
Induced EMF is in direction that opposes
the change in flux that caused it

9. P22- 9
Faraday’s Law of Induction
Bd
dt
ε Φ
= −
Changing magnetic flux induces an EMF
Lenz: Induction opposes change

10. P22-10
PRS: Faraday’s Law: Loop
A coil moves up
from underneath a
magnet with its
north pole pointing
upward. The
current in the coil
and the force on the
coil:
0%
0%
0%
0%
:00
1. Current clockwise; force up
2. Current counterclockwise; force up
3. Current clockwise; force down
4. Current counterclockwise; force down

11. P22-11
PRS Answer: Faraday’s Law: Loop
The I dl x B force on the coil is a force which is
trying to keep the flux through the coil from
increasing by slowing it down (Lenz’s Law again).
Answer: 3. Current is clockwise; force is down
The clockwise current
creates a self-field
downward, trying to offset
the increase of magnetic
flux through the coil as it
moves upward into stronger
fields (Lenz’s Law).

12. P22-
Changing Angle
B BAΦ = × =B A
rr
0BΦ = × =B A
rr

13. P22-13
Part 1: Current & Flux
Current?
Flux?
I > 0
BLACK
RED
( )
0
( ) ' '
t
t R I t dtΦ = − ∫

14. P22-14
PRS: Flux Measurement
Moving from above to below and back, you will
measure a flux of:
t t
t t
(A)
(C)
(
B)
(D)
1
2
3
4
5
6
7
8
0% 0% 0% 0%0%0%0%0%
1. A then A 5. B then B
2. C then C 6. D then D
3. A then C 7. B then D
4. C then A 8. D then B5. 5
6. 6
7. 7
8. 8
0

15. P22-15
Self Inductance
11 11 1M I LIΦ = Φ ≡ ≡
dI
L
dt
ε = −
What if we forget about coil 2 and
ask about putting current into coil 1?
There is “self flux”:
Faraday’s Law 

16. P22-16
Inductor Behavior
I
Inductor with constant current does nothing
dI
L
dt
ε = −
L

17. P22-17
1. Start with “uncharged” inductor
2. Gradually increase current. Must work:
3. Integrate up to find total work done:
Energy To “Charge” Inductor
dI
dW Pdt I dt L I dt LI dI
dt
ε= = = =
21
2
0
I
I
W dW LI dI L I
=
= = =∫ ∫

18. P22-18
ANY QUERY
AND
THANKYOU