Electrodynamic, Electromagnetic induction

1. Gaurav Kumar Yogesh
413115003
Department of Physics
NIT –Trichy 620015

2.  In electrodynamics, changing electric and magnetic field with time.
 Fields are time variables.
 Static magnetic field produced by steady current and static electric
field produced by the charge at rest.
 If the current is time dependent the magnetic field is also time
dependent.
 Thus, if magnetic field is dependent then electric is also time
dependent.
)()()( tEtBtI 

3.  Electromotive force is work done
per unit charge.
 If we have a wire connected to a
battery then it is conducting and
battery supply the force then the
free charge in a wire are in motion
and produce work.
 dlE.

4.  Only the part AB will acts as a battery.
And current will flow from A to B.
 If the direction of motion of closed loop is
opposite then direction of flow of current is
also get opposite
dt
d
dsBm




 .
 dsfmag .
)(
)(
veVBf
veBVf
mag
mag

 If charge is positive
If charge is negative
R
VBh
R
v
iiRv
VBh
dlVB
dlfmag








 .

5. Ans. Due to motion of the motion of the disc the charge in
the motion. So the magnetic force per unit charge is
Because of this force, the charge moves toward the outer
edge. So outer edge of the terminal becomes the +ve
charge and centre becomes –ve charge.
If the disc rotates oppositely, the direction of force is
opposite. Then the charge on the centre is +ve and
surface is –ve.
Question: A metal disc of radius ‘a’ rotates with angular velocity w about vertical
axis through a uniform magnetic field B pointing in the upward direction. A circuit
is made By connecting to one end of resistor to the axle and other to the sliding
contact which touches the outer edge of the disc. Find the current in the resistance
in the resistor with direction




ˆ
)ˆˆ(
ˆ
rV
rzrV
rV
z




rrBf
zrBf
zBrf
BVf
mag
mag
mag
mag
ˆ
)ˆˆ(
ˆ







2
.
2
0
Ba
drrB
a




 

6. Ans: (a) The potential is
Inside the rod the current flow from B to A
(b) The magnetic force is
And the force is acting toward the left side of the bar.
Velocity of bar is exponentially decreases with time.
Question: A metal bar of mass m slides frictionally o two parallel conduction rails. A
distance l apart a resistor R is connected across the rails and a uniform mag. Field B is
pointing into the page
(a) Find the magnitude of current and magnetic force on the bar.
(b) If V0 is the speed at the t = 0, then what is the speed at later time t.
R
VBl
iVBl
dlfmag

 

 .
R
lVB
lB
R
VBl
f
BdlIf
mag
mag
22
. 
 
t
mR
lB
o
o
eVV
t
mR
lB
V
V
Ct
mR
lB
V
dt
mR
lB
V
dV
mR
lVB
dt
dV
R
lVB
dt
dV
m
22
22
22
22
2222
)ln(
ln







7. Ans : square loop is pulled upward direction.
Magnetic field due to a wire is
Where s is variable, if the s is increases the flux trough the wire is decreases.
Question: A square loop of wire of side ‘a’ lies on a table at a distance s from a very
long straight wire carries a current I. if someone pulls the loop away from the wire at
speed V. What is the induced emf in the loop and also find the induced current.
)ln(
2
2
ˆ
2
,0
s
as
a
I
dxdy
s
I
dxdyds
s
I
B
o
m
assy
ax
o
m
o

















dt
ds
s
a
as
s
a
I
dt
d
o
m
))((
2 2










8.  As the loop is moving away from the wire (=s) here the flux is decreases,
so the induced current is try to induced it and flux is decreasing due to the
decrease in the magnetic field.
 Field due to the induced current will try to increase the magnetic field , to
do so., it will develop magnetic field opposite to the original.
 If we pull the loop toward the ire the flux and B is increase, so current try to
decrease it.
 If we pull the loop away from the wire. The induced current in such a way
that it increase the current in wire.
 If the loop run parallel to the wire the then no emf will generate to the wire.

9. If the loop is state of rest and B is changing.
Then EMF is
so we have induced electric field only if we
have B is changing with time. Here E is not
a conservative force because curl of E in not
equal to zero.
E- field lines – open curves
B- field lines – closed curves
Field lines in the closed curves due to induced
electric field is
 Properties of induced electric firld
1. Non conservative
2. Non electrostatic
3. Closed Curve
4. is because there is no charge
corresponds to it otherwise
If both loop and magnetic field is changing
dt
Bd
E
dsB
dt
d
E
dsEdsB
dt
d
dlEdsB
dt
d
dlE
dt
d
dlE
dlE
m













0).)((
).(.
..
.
0.
.





0.  E
0.  E
o
E


.






dlBVdsB
dt
d
dlfdsB
dt
d
dlEdsB
dt
d
mag
.)(.
..
..




10. Ans: we have
B is not function of distance i.e., B is uniform. So flux is
Direction of current is changing accordingly Sin(wt).
Question: A long solenoid of radius ‘a’ is driven by an alternating current such that
field inside the solenoid is . A circular loop of wire or radius “a/2”
and resistance R is placed inside the solenoid and coaxial with it. Find the current
induced in the loop as function of time
ztCosBtB o ˆ)()( 
ztCosBtB o ˆ)()( 
R
tSin
a
B
R
i
tSin
a
B
dt
d
tCos
a
B
a
B
o
o
m
om
m
)()
4
(
)()
4
(
)()
4
(
)
4
.(
2
2
2
2


















11.  We have
Flux is changing with time, the emf of the loop is obtained as
The current producing such that the magnetic field is clockwise.
Note: Induced current always less than the original current
except the case of superconductor
Question: A square loop of wire with side ‘a’ lies in the 1st quadrant of x-y plane with
one corner at the origin. In this region there is non-uniform time dependent magnetic field
is coming out of the page. Where k is some constant. Find the induced
EMF in the loop.
ztkytyB ˆ),( 23

44
.
.
ˆ),(
5
2
4
2
0
,0
23
23
a
kt
a
akt
dxdytky
dsB
ztkytyB
m
ay
ax
m
m











ztkytyB ˆ),( 23

2
)
4
(
5
5
2
kta
a
kt
dt
d





12. Ans:
The end becomes the Positively charged at the end A.
Question: A metallic rod AB with uniform velocity v in a uniform magnetic field B as
shown in figure.
(a) The rod become electrically charged
(b) (b) The end A becomes +ve charged
(c) The end B becomes +ve charged
(d) The rod becomes hot because of the joule heating

13. Ans: we know that rod having the angular momenta in z
Direction
So answer (b) is the correct option
The potential difference between the A and B is zero.
A rod of length ‘a’ rotates with a uniform angular velocity ‘w’ about its perpendicular
bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential
Difference between the centre of the rod and end is
(a) 0
(b) 1/8(B.a.a.w)
(c) ½(wB.a. a)
(d) Bl2w
8
.
ˆ
)ˆˆ(
ˆ
2
2/
0
a
B
rdrB
dlf
rV
rzrV
z
ar
r
mag

















14.  Since the copper tube can be understand as circular wire, so the flux is change
in the motion of the magnet. As the magnet passes it emf such that, it ca resist
the motion of magnet at the first end of the tube and it will induce the emf at
the second end such that induce motion also resist the motion of the magnet
and the motion inside the tube is almost constant.
 So the magnet move with constant velocity
Question: A bar is released from the rest along the axis of a very long vertical copper tube.
After some time the magnet:
(1) Will stop in the tube
(2) Will move with almost constant speed
(3) Will move with an acceleration g.
(4) Will oscillate

15. Ans: Induced electric field
If
1. Here B is increases wit t, as flux also increases. S the induce current or induce
field try to decrease the flux. So it will opposite to actual current. So the
direction of the induce current is clockwise
2. If B is decrease with time the induce current is opposite of the previous case.
Question: A uniform magnetic field B(t) pointing in the direction fills the circular region
as shown in the figure. If B is changing with time. what is induced electric field.
dt
tdBs
E
s
dt
tdB
SE
s
dt
tdB
dlE
stB
dsB
dt
d
dlE
m
m
m
)(
2
.
)(
2.
.
)(
.
).(
.
.
2
2
2














zatB ˆ2


16.  Bo is constant not depending on time
Then flux is
Now B is switched off i.e., B = 0, & flux = 0,
Force required to rotate the charge, electric
field will do work for rotation of charge
electric force
Question: A line charge lambda is uniformly distributed on the rim of a wheel of a radius
B which is then suspended horizontally as shown in the figure & and it is free to rotate.
In the central region up to a radius ‘a’ there is a uniform magnetic field B pointing in the
Up direction . Suppose someone turned off the magnetic field, then what is the angular
velocity in the wheel.
2
. aBm  
Frtorque
dt
dB
a
dt
d
dlE
o
m



2
.


)(
)(
.
.
2
2
tdBbaLd
dt
tdB
ba
dt
Ld
N
dlEbN
dlEF
Ef












When B= Bo, L=0 & B= 0, L=L;
So,
And mvrL
BbaL
dBbadL
oB
L


 


2
0
2
0
mb
Ba 

2

B

17. Ans: So, Magnetic field inside
the solenoid is
 If B is in Z direction & Ein
(induced) will be in phi
direction
Question: A long solenoid with radius ‘a’ & n turns per unit length carries a time
dependent current I(t) in the direction. Find the electric field both direction and
magnitude at distance s from the axis both inside and outside the solenoid.











ˆ)(
2
)(
ˆ)(
)(2.
))((
ˆ)(
2
)(
ˆ)(
)(2.
ˆ)(
)(.
))((
)(
.
2
2
2
2
2
2
dt
tdI
s
an
E
dt
tdI
ansE
atIn
dt
tdIsn
E
dt
tdI
snsE
dt
tdI
sndlE
stIn
tInB
dt
d
dlE
o
o
om
o
o
o
om
o
m










 Outside: Flux =
Although the magnetic flux outside the
Solenoid is zero, but induced magnetic
Field is not zero
If current is increases with time then induced
current oppose the flux, if current decreases with
time then create the flux.




ˆ)(
2
)(
ˆ)(
)(2.
))((
2
2
2
dt
tdI
s
an
E
dt
tdI
ansE
atIn
o
o
om



s
EsE outin
1
, 
ˆ

18. Ans: we know that
Question: A square loop of side a & resistance R lies at a distance s from an infinite
Straight wire that carries a current I. Now someone cut the wire such that I drops to
Zero. In what direction induce current in loop will flow. And calculate the total charge
Passes a given point in the loop during this time of current flow.
)ln(
2
)ln(
2
)ln(
2
)ln(
2
)ln(
2
.
)ln(
2
)]ln(
2
[
2
ˆ
2
.
0
s
as
R
Ia
Q
dI
s
as
R
a
dQ
dt
dI
s
as
R
a
dt
dQ
dt
dQI
dt
dI
s
as
R
a
I
dt
dI
s
asa
RI
dt
dI
s
asIa
s
asIa
dt
d
dy
s
Ia
dt
d
dxdyds
s
I
B
dt
d
dlE
o
I
o
o
o
loop
o
loop
o
o
ass
as
o
o
m














































 Direction of the induced current– Anticlockwise
If current decreases, the flux decreases so the
Induced will try to increase the flux.

19. Ans: Area is fixed, B is also fixed but the angles changes.
Question: A square loop of edge a having ‘n’ turns is rotated with a uniform angular
Velocity ‘w’ about one of its diagonal in horizontal position as shown in this figure.
Find the induced EMF in the coil and also plot the graph between induced current and
phase with respect.
R
tSinBa
R
i
tSinBa
dt
d
tCosBa
CosBa
m
m
m
)(
)(
)(
2
2
2
2











20. Any Queries
???