Download DC electrical circuits

DC Electrical Circuits
Chapter 28
(Continued)
Circuits with Capacitors
Kirchhoff’s Laws
The loop method is based on two laws devised
by Kirchoff:
1. At any circuit junction,
currents entering must
equal currents leaving.
2. Sum of all DV’s across
all circuit elements
in a loop must be zero.
I2
I1
I3= I1+ I2
I
r
E+
-
R
E - Ir - IR = 0
RC Circuits: Charging
open
closed
R
+
E
-
C
I
+
E
-
R
VR=IR
+++
---
C VC=q/C
When the switch closes, at first a high current flows:
VR is big and VC is small.
RC Circuits: Charging
open
closed
R
+
E
-
C
I
+
E
-
R
VR=IR
+++
---
C VC=q/C
When the switch closes, at first a high current flows:
VR is big and VC is small.
As q is stored in C, VC increases.
This fights against the battery so I decreases.
RC Circuits: Charging
Apply the loop law: E – IR - q/C = 0
closed
I
+
E
-
R
Take the derivative of this
with respect to time:
VR=IR
dI 1 dq
R

0
dt C dt
+++
---
C
VC=q/C
Now use dq/dt = I 
and rearrange:
dI
I

dt
RC
This is a differential equation for an unknown function I(t).
It is solved subject to the initial condition I(0) = E / R.

RC Circuits: Charging
dI
I  dI
dt


dt
RC
I
RC
closed
I
+
E
-
R
VR=IR
+++
---

C
VC=q/C
dI
dt
 I    RC
 t
ln I  
b
RC
I(t)  I0e
And I(0) = I0 = E / R 

I
 t RC
E
R
where I0  e b
t / RC
e
RC Circuits: Charging
closed
I
R
+
E
-
VR=IR
+++
---
C
From this we get:
VC=q/C
I
E
R
t / RC
e
t / RC
E
VR  IR  e
t / RC
V

)
C E VR  E (1 e
q = VC C = E C (1 – e–t/RC)


Charging
Potential Drop
Current
E/R
I
t/RC
E
VC
VR
t/RC
Discharging an RC Circuit
C
R
R
q
q
-q
VC=V0
Open circuit
I
C
VR=IR
-q VC=q/C
After closing switch
Current will flow through the resistor for a while.
Eventually, the capacitor will lose all its charge, and the
current will go to zero.
Power P = IV = I2R will be dissipated in the resistor
(as heat) while the current flows.
Discharging an RC Circuit
R
I
C
Loop equation:
q/C - IR = 0  I = q / (RC)
VR=IR
+q
-q VC=q/C
dI
I

Take d/dt 
dt
RC
[Note that I = - dq/dt]
 the initial voltage on the capacitor:
Here the current at t=0 is given by
I(0) = V0/R = q0 /RC
This equation is solved very much like the other (charging case):
V0 t / RC
I
e
R
Discharging an RC Circuit
R
I
C
VR=IR
V0 t / RC
I e
R
q
-q
VC=q/C
The charge 
on the capacitor is given by:
q/C - IR = 0 so q = C IR [q = C V]
q  CV0 e
 q0 e
t / RC
t / RC
Discharging
Current
E/R
Potential Drop
t/RC
VR
0
VC
t/RC

Example: A capacitor C discharges through a resistor R.
(a) When does its charge fall to half its initial value ?
Charge on a capacitor varies as
Q  Q0 e
t / RC
R
Q
I
C

Example: A capacitor C discharges through a resistor R.
(a) When does its charge fall to half its initial value ?
Charge on a capacitor varies as
Q  Q0 e
t / RC
Find the time for which Q=Q0/2
R
Q
I
C

Example: A capacitor C discharges through a resistor R.
(a) When does its charge fall to half its initial value ?
Charge on a capacitor varies as
Q  Q0 e
t / RC
Find the time for which Q=Q0/2
1
t / RC
Q0  Q0e
2
t
ln 2  
RC
t  (ln 2)RC  0.69RC
R
Q
I
C

Example: A capacitor C discharges through a resistor R.
(a) When does its charge fall to half its initial value ?
Charge on a capacitor varies as
Q  Q0 e
t / RC
Find the time for which Q=Q0/2
1
t / RC
Q0  Q0e
2
t
ln 2  
RC
t  (ln 2)RC  0.69RC
R
Q
I
C
RC is the
“time constant”
Example: A capacitor C discharges through a resistor R.
(b) When does the energy drop to half its initial value?
The energy stored in a capacitor is
2
2
0
Q Q 2t / RC
U(t)   e
 U 0e2t / RC
2C 2C
We seek the time for U to drop to U0/2:
1
U 0  U 0e2t / RC
2
ln 2
 t  ln 2  RC
 0.35RC
2
Example: A capacitor C discharges through a resistor R.
(b) When does the energy drop to half its initial value?
The energy stored in a capacitor is
2
2
0
Q Q 2t / RC
U(t)   e
 U 0e2t / RC
2C 2C
We seek the time for U to drop to U0/2:
1
U 0  U 0e2t / RC
2
ln 2
 t  ln 2  RC
 0.35RC
2
Magnetic Fields
Chapter 29
Permanent Magnets & Magnetic Field Lines
The Magnetic Force on Charges
Magnetism
• Our most familiar experience of magnetism is through permanent
magnets.
• These are made of materials which exhibit a property called
ferromagnetism - i.e., they can be magnetized.
• Depending on how we position two magnets, they will attract or repel,
i.e. they exert forces on each other.
• Just as it was convenient to use electric fields instead of electric forces,
here too it is useful to introduce the concept of the magnetic field B.
• There are useful analogies between electric and magnetic fields, but
the analogy is not perfect: while there are magnetic dipoles in nature,
there seem to be no isolated magnetic charges (called “magnetic
monopoles”). And the force laws are different.
• We describe magnets as having two magnetic poles:
North (N) and South (S).
• Like poles repel, opposite poles attract.
Field of a Permanent Magnet

B
N
S
Shown here are field lines. The magnetic field B at
any point is tangential to the field line there.
Field of a Permanent Magnet

B
N
N
S
S
The south pole of the small bar magnet is attracted towards
the north pole of the big magnet.
Also, the small bar magnet (a magnetic dipole) wants to align
with the B-field.
The field attracts and exerts a torque on the small magnet.
Magnetism
• The origin of magnetism lies in moving electric charges.
Moving (or rotating) charges generate magnetic fields.
• An electric current generates a magnetic field.
• A magnetic field will exert a force on a moving charge.
• A magnetic field will exert a force on a conductor that
carries an electric current.
What Force Does a Magnetic Field Exert
on Charges?

B
q
• If the charge is not moving with
respect to the field (or if the
charge moves parallel to the
field), there is NO FORCE.
What Force Does a Magnetic Field Exert
on Charges?

B
• If the charge is not moving with
respect to the field (or if the
charge moves parallel to the
field), there is NO FORCE.
q
v
q
B
• If the charge is moving, there
is a force on the charge,
perpendicular to both v and B.
F=qvxB
Force on a Charge in a
Magnetic Field
F  qv  B
or
F  qvBsin 
(Use “Right-Hand” Rule to
determine direction of F)
F
v
B
mq
Units of Magnetic Field
Since
F  qvBsin 
F
B
qvsin 
Therefore the units of magnetic
field
 are: N  Ns/Cm 1T (Tesla)
C  m /s

(Note: 1 Tesla = 10,000 Gauss)
The Electric and Magnetic Forces are
Different
Whereas the electric force
acts in the same direction as
the field:
The magnetic force acts in a
direction orthogonal to the
field:


F  qE
F  qv  B
The Electric and Magnetic Forces are
Different
Whereas the electric force
acts in the same direction as
the field:
F  qE
The magnetic force acts in a
direction orthogonal to the
field:


F  qv  B
(Use “Right-Hand” Rule to
determine direction of F)
The Electric and Magnetic Forces are
Different
Whereas the electric force
acts in the same direction as
the field:
The magnetic force acts in a
direction orthogonal to the
field:

F  qE
F  qv  B
(Use “Right-Hand” Rule to
determine direction of F)
And – the charge must be moving.

Trajectory of Charged Particles
in a Magnetic Field
(B field points into plane of paper.)
+
B
+
+
+
+
+
+
+
+ F
+
+
+
+
+
+
+
+
+
+
v
+
Trajectory of Charged Particles
in a Magnetic Field
(B field points into plane of paper.)
+
+B
+
v+
+
+
+
+
+
+
+
+ F
+
+
+
+ F +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
B
+
+
+
+
+
v
Trajectory of Charged Particles
in a Magnetic Field
(B field points into plane of paper.)
+
+B
+
v+
+
+
+
+
+
+
+
+ F
+
+
+
+ F +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
B
+
+
+
+
+
v
Magnetic Force is a centripetal force
Rotational Motion

 = s / r  s =  r  ds/dt = d/dt r  v =  r
s
r
 = angle,  = angular speed,  = angular acceleration
at = r 
tangential acceleration
ar = v2 / r radial acceleration

at
ar
The radial acceleration changes the direction of motion,
while the tangential acceleration changes the speed.
Uniform Circular Motion
ar
 = constant  v and ar constant but direction changes

v
ar = v2/r = 2 r
KE = (1/2) mv2 = (1/2) m2r2
F = mar = mv2/r = m2r
Radius of a Charged Particle
Orbit in a Magnetic Field
+B
+
v+
+
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
+
F
+
Centripetal
Force
=
Magnetic
Force
Radius of a Charged Particle
Orbit in a Magnetic Field
+B
+
+
v+
+
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+ 
F
+
Centripetal
Force
mv

r
=
2
Magnetic
Force
 qvB
Radius of a Charged Particle
Orbit in a Magnetic Field
+B
+
+
v+
+
Centripetal
Force
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
mv 2

 qvB
r
F
+
=
Magnetic
Force


mv
r
qB
Radius of a Charged Particle
Orbit in a Magnetic Field
+B
+
+
v+
+
Centripetal
Force
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
mv 2

 qvB
r
F
+
=
Magnetic
Force


mv
r
qB
Note: as Fv , the magnetic
force does no work.
Cyclotron Frequency
+B
+
v+
+
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
+
The time taken to complete one
orbit is:
2r
T
v
2 mv

v qB
F
+

Cyclotron Frequency
+B
+
v+
+
+
+
+
+
+
+
+
r
+
+
+
The time taken to complete one
orbit is:
2r
T
v
2 mv

v qB
F
+
+
+
+
+
+
Hence the orbit frequency, f

1
qB
f  
T 2 m
Cyclotron Frequency
+B
+
v+
+
+
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
F
+
The time taken to complete one
orbit is:
2r
T
v
2 mv

v qB
Hence the orbit frequency, f
1
qB
f  
T 2 m
- known as the “cyclotron frequency”
T = 2/ = 1/ƒ  ƒ = /2
The Electromagnetic Force
If a magnetic field and an electric field are simultaneously
present, their forces obey the superposition principle and
must be added vectorially:

B+

+
+
+
+
F  qE  qv  B

+v +
+
q
+
+
+
+
+
+
+

E
The Lorentz force
Here FE and FB point
in opposite directions
Exercise
electron
v
B
v’
• In what direction does the magnetic field point?
• Which is bigger, v or v’ ?
Exercise: answer
electron
v
B
F
v’
• In what direction does the magnetic field point ?
Into the page [F = -e v x B]
• Which is bigger, v or v’ ?
v = v’ [B does no work on the electron, Fv]
What is the orbital radius of a charged particle (charge q, mass m)
having kinetic energy K, and moving at right angles to a magnetic
field B, as shown below?.
x x x
B
x x x
K
q• m
What is the orbital radius of a charged particle (charge q, mass m)
having kinetic energy K, and moving at right angles to a magnetic
field B, as shown below?.
F = q v x B = m a and a = v2 / r
q v B = m v2 / r
x x x
B
x x x
qB=mv/rrqB=mv
r
K = (1/2) mv2
q• m
r = m v / (q B)
r2 = m2 v2 / (q B)2
(1/2m) r2 = K / (q B)2  r = [2mK]1/2 / (q B)
What is the relation between the intensities of the electric and
magnetic fields for the particle to move in a straight line ?
x x x B
E
x x x
v
q• m
The magnetic field points into the
picture. The direction of the electric
field is not yet specified.
What is the relation between the intensities of the electric and
magnetic fields for the particle to move in a straight line ?
x x x B
E
x x x
v
q• m
FE = q E and FB = q v B
If FE = FB the particle will move
following a straight line trajectory
qE=qvB
FB FE
•
v=E/B
What is the relation between the intensities of the electric and
magnetic fields for the particle to move in a straight line ?.
x x x B
E
x x x
v
q• m
FE = q E and FB = q v B
If FE = FB the particle will move
following a straight line trajectory
qE=qvB
FB FE
•
v=E/B
So need E pointing to the right.
Trajectory of Charged Particles
in a Magnetic Field
What if the charged particle has a velocity
component along B?


Vz unchanged
Vz

B
Circular motion in
xy plane.
x
z
y