ELECTROMAGNETIC INDUCTION

1. Earlier we had defined the flux of any vector field. For a
magnetic field, this means that flux of a uniform magnetic
field (see figure) is cos . If the field is not
constant ove
B A BA θ ⊥ Φ = =
2
r the area then we must add up all the little
pieces of flux: . The dimension of flux is
magnetic field area, and the unit is called weber, where
1 weber 1 tesla metre .
ΦB = B⋅dA
×
= ⋅
∫
􀁇 􀁇
2. A fundamental law of magnetism states that the net flux through a closed surface is always
zero, 0. Note that this is very different from what you learned earlier in
electrostati
B Φ = ∫B⋅dA=
􀁇 􀁇
cs where the flux is essentially the electric charge. There is no such thing as a
magnetic charge! What we call the magnetic north (or south) pole of a magnet are actually
due to the particular electronic currents, not magnetic charges. In the bar magnet below,
no matter which closed surface you draw, the amount of flux leaving the surface is equal
to that entering it.
A sphere of radius is placed near a long, straight wire that carries a steady
current . The magnetic field generated by the current is . Find the total magnetic flux
passing thro
R
I B
Example :
ugh the sphere.
B 􀁇
Answer: zero, of course!
3. Faraday's Law for Induced EMF: when the magnetic flux changes in a circuit, an electromotive
force is induced which is proportional to the rate of change of flux. Mathematically,
where is the induced emf. If the coil consists of turns, then .
How does th
B B d N Nd
dt dt
ε Φ ε ε Φ
= − = −
e flux through a coil change? Consider a coil and magnet. We can:
a) move the magnet,
b) change the size and shape of the coil by squeezing it,
c) move the coil.
In all cases, the flux through the coil changes and is non-zero leading to an induced
emf.
: A flexible loop has a radius of 12cm and is in a magnetic field of strengt
B d
dt
Φ
Example h 0.15T.
The loop is grasped at points A and B and stretched until it closes. If it takes 0.20s to close
the loop, find the magnitude of the average induced emf in it during this time.
Solu
2
tion: Here the loop area changes, hence the flux. So the induced emf is:
final flux initial flux 0 (0.12) 0.15 0.034 Volts.
time taken 0.2
B d
dt
ε = − Φ ≈ −⎢⎣⎡ − ⎥⎦⎤= −⎢⎣⎡ −π × ⎥⎦⎤=
A wire loop of radius 0.30m lies so that an
external magnetic field of +0.30T is perpendicular to
the loop. The field changes to -0.20T in 1.5s. Find the
magnitude of the aver
Example :
2
age induced EMF in the loop
during this time.
Solution: Again, we will find the initial and final fluxes
first and then divide by the time taken for the change.
Use Φ = BA = Bπ r to calcul
2 2
2 2
ate the flux.
0.30 (0.30) 0.085 Tm
0.20 (0.30) 0.057 Tm
0.085 0.057 0.095 V
1.5
i
f
f i
t t
π
π
ε
Φ = × × =
Φ = − × × = −
ΔΦ Φ −Φ −
= = = =
Δ Δ
4. Remember that the electromotive force is not really a force but the difference in electric
potentials between two points. In going around a circular wire, where the electric field is
constant as a function of angle, the emf is (2 ). More generally, for any size or
shape of a closed circuit, . So Faraday's Law reads: B .
E r
E ds E ds d
dt
ε π
ε
=
Φ
=∫ ⋅ ∫ ⋅ =−
􀁇 􀁇 􀁇 􀁇
A conducting wire rests upon two parallel rails and is pulled towards the right
with speed v. A magnetic field B is perpendicular to the plain of the rails as shown below.
Find the
Example :
current that flows in the circuit.
Solution: As the wire is pulled to the right, the area of the circuit increases and so the
flux increases. Measure as above so that v , i.e. keeps increasing as we pull.
The fl
t
ux a a
x x= t x
( )
ny value of is , and so,
v.
To calculate the current, we simply use Ohm's Law: v . Here is the
resistance of the circuit.
B
B
x BDx
d d BDx BDdx BD
dt dt dt
I BD R
R R
ε
ε
Φ =
Φ
= − = − = − =
= =
2
2 2 2
2
Find the power dissipated in the above circuit using , and then by directly
calculating the work you do by pulling the wire.
Solution: Clearly v is the power
I R
I R B D
R
=
Example :
dissipated, as per usual formula. Now let us
calculate the force acting upon the piece of wire (of length ) that you are pulling. From
the formula for the force on a wire, , the magn
D
F=IL×B
􀁇 􀁇 􀁇 2 2
2 2 2
itude is v . So,
the power is v v . This is exactly the value calculated above!
F IBD B D
R
P F B D
R
= =
= =
x
v 􀁇
2 F
􀁇
3 F
􀁇
(into paper) B
􀁇
D
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
⊗
5. The direction of any magnetic induction effect is such as to oppose the cause
of the effect. Imagine a coil wound with wire of finite resistance. If the magnetic field
decreases
Lenz's Law :
, the induced EMF is positive. This produces a positive current. The magnetic
field produced by the current opposes the decrease in flux. Of course, because of finite
resistance in loop, the induced current cannot completely oppose the change in flux.