DC Electrical Circuits

DC Electrical CircuitsDC Electrical Circuits

Chapter 28Chapter 28(Continued)

Circuits with CapacitorsCircuits with Capacitors

Kirchhoff’s Laws

1. At any circuit junction, currents entering must equal currents leaving.

The loop method is based on two laws devised by Kirchoff:

2. Sum of all V’s across all circuit elements in a loop must be zero.

I1

I2

I3= I1+ I2

+-

r

ER

I

E - Ir - IR = 0

+++- - -

RC Circuits: Charging

VR=IR

VC=q/C

When the switch closes, at first a high current flows:VR is big and VC is small.

+

R

C

open closed

E

I

+E

R

C- -

+++- - -


VR=IR

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.

+

R

C

open closed

E

I

+E

R

C- -


Now use dq/dt = I and rearrange:

Apply the loop law: E – IR - q/C = 0

Take the derivative of thiswith respect to time:

€

dIdt

= − IRC

This is a differential equation for an unknown function I(t).It is solved subject to the initial condition I(0) = E / R.

€

−R dIdt

− 1C

dqdt

= 0

VR=IR

VC=q/C+

R

C

closed

E

I

- - - -

+++

€

I =R

e−t / RC


€

dIdt

= − IRC

€

dII

= − dtRC

And I(0) = I0 = E / R E

VR=IR

VC=q/C+

R

C

closed

E

I

- - - -

+++

€

dII∫ = − dt

RC∫

ln I = − tRC

+ b

I(t) = I0e− t

RC where I0 = eb

€

(1− e−t / RC )


From this we get:

€

VR = IR = E

€

e−t / RC

VC =E

€

−VR =Eq = VC C = E C (1 – e–t/RC)

VR=IR

VC=q/C+

R

C

closed

E

I

- - - -

+++

€

I =R

e−t / RCE

t/RC

t/RC

Pote

ntia

l Dro

p

E

E/R

Cur

rent

VC

VR

Charging

I

Discharging an RC Circuit

R

Cq

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.

-q

Open circuit

VC=V0

R

CqI-q

After closing switch

VR=IR

VC=q/C


R

C+qI -q

VR=IR

VC=q/C

Loop equation:q/C - IR = 0 I = q / (RC)

Take d/dt

€

dIdt

= − IRC

Here the current at t=0 is given by the initial voltage on the capacitor: I(0) = V0/R = q0 /RC

€

I = V0

Re−t / RC

[Note that I = - dq/dt]

This equation is solved very much like the other (charging case):


R

CqI-q

The charge on the capacitor is given by: q/C - IR = 0 so q = C IR [q = C V]

€

q = CV0 e−t / RC

€

=q0 e−t / RC

VR=IR

VC=q/C

€

I = V0

Re−t / RC

t/RC

t/RC

Pote

ntia

l Dro

p

0

E/R

Cur

rent

VC

VR

Discharging

Example: A capacitor C discharges through a resistor R.(a) When does its charge fall to half its initial value ?

R

CQ

I

Charge on a capacitor varies as

€

Q = Q0 e−t / RC


RCharge on a capacitor varies as

€

Q = Q0 e−t / RC

Find the time for which Q=Q0/2 C

Q

I



€

Q = Q0 e−t / RC

Find the time for which Q=Q0/2

€

12

Q0 = Q0e−t / RC

∴ −ln2 = − tRC

t = (ln2)RC = 0.69RC

CQ

I



Find the time for which Q=Q0/2

€

Q = Q0 e−t / RC

RC is the “time constant”

CQ

I

€

12

Q0 = Q0e−t / RC

∴ −ln2 = − tRC

t = (ln2)RC = 0.69RC

€

U(t) = Q2

2C= Q0

2

2Ce−2t / RC = U0e

−2t / RC

12

U0 = U0e−2t / RC

∴ t = −ln2 × RC ln22

= 0.35RC

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

We seek the time for U to drop to U0/2:

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

We seek the time for U to drop to U0/2:

€

U(t) = Q2

2C= Q0

2

2Ce−2t / RC = U0e

−2t / RC

12

U0 = U0e−2t / RC

∴ t = −ln2 × RC ln22

= 0.35RC

Magnetic FieldsMagnetic FieldsChapter 29Chapter 29

Permanent Magnets & Magnetic Field LinesPermanent Magnets & Magnetic Field LinesThe Magnetic Force on ChargesThe 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

N S

rB

Shown here are field lines. The magnetic field B at any point is tangential to the field line there.

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.

Field of a Permanent Magnet

N S

rB

N

S

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?

• If the charge is not moving with respect to the field (or if the charge moves parallel to the field), there is NO FORCE.

rB

q

What Force Does a Magnetic Field Exert on Charges?

rB

q

€

r B q

€

r v

• If the charge is moving, thereis a force on the charge,perpendicular to both v and B.

F = q v x 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.

Force on a Charge in aMagnetic Field

€

r F = q

r v ×

r B

orF = qvBsinθ

vF

Bqm

(Use “Right-Hand” Rule to determine direction of F)

Units of Magnetic Field

Since

€

F = qvBsinθ

€

B = Fqvsinθ

Therefore the units of magnetic field are:

€

NC ⋅m /s

= Ns /Cm =1T (Tesla)

(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:

€

r F = q

r v ×

r B

€

r F = q

r E




€

r F = q

r v ×

r B

€

r F = q

r E




And – the charge must be moving.


€

r F = q

r v ×

r B

€

r F = q

r E


Trajectory of Charged Particlesin a Magnetic Field

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

(B field points into plane of paper.)


+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vvB B

F F



+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vvB B

F F


Magnetic Force is a centripetal force

Rotational Motion

r s = s / r s = r ds/dt = d/dt r v = r

atar

at = r tangential accelerationar = v2 / r radial acceleration

The radial acceleration changes the direction of motion,while the tangential acceleration changes the speed.

Uniform Circular Motion

= constant v and ar constant but direction changes

ar = v2/r = 2 rF = mar = mv2/r = m2r

KE = (1/2) mv2 = (1/2) m2r2

= angle, = angular speed, = angular acceleration

v

ar

Radius of a Charged ParticleOrbit in a Magnetic Field

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

r

Centripetal Magnetic Force Force=


€

∴mv2

r= qvB


+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

r


€

∴mv2

r= qvB

⇒ r = mvqB


+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

r

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +


vB

F

r

€

∴mv2

r= qvB

⇒ r = mvqB

Centripetal Magnetic Force Force =

Note: as , the magneticforce does no work.

€

r F ⊥r

v

Cyclotron Frequency

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

r

The time taken to complete one orbit is:

€

T = 2πrv

= 2πv

mvqB

Cyclotron Frequency

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

r


Hence the orbit frequency, f

€

f = 1T

= qB2π m€

T = 2πrv

= 2πv

mvqB

Cyclotron Frequency

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

r


T r

mqB

=

=

2

2

π

πv

vv

Hence the orbit frequency, f

- known as the “cyclotron frequency”

f =1T

=B2π m

T = 2π/ = 1/ƒ ƒ = /2π

The Electromagnetic Force

If a magnetic field and an electric field are simultaneouslypresent, their forces obey the superposition principle and must be added vectorially:

+ + ++ + ++ + ++ + ++ + +

rvrB r

Eq

€

Herer F E and

r F B point

in opposite directions

€

r F = q

r E + q

r v ×

r B The Lorentz force

Exercise

Bv

v’

• In what direction does the magnetic field point?

• Which is bigger, v or v’ ?

electron

Exercise: answer

Bv

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, Fv]

electron

F

What is the orbital radius of a charged particle (charge q, mass m) having kinetic energy K, and moving at right angles to a magneticfield B, as shown below?.

xx xx

xx

•q m

B

K

What is the orbital radius of a charged particle (charge q, mass m) having kinetic energy K, and moving at right angles to a magneticfield B, as shown below?.

xx xx

xx

•q m

Br

K = (1/2) mv2

q v B = m v2 / r

F = q v x B = m a and a = v2 / r

q B = m v / r r q B = m v

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 ?

The magnetic field points into the picture. The direction of the electric field is not yet specified.

xx xx

xx

•q m

B

v

E

What is the relation between the intensities of the electric and magnetic fields for the particle to move in a straight line ?

xx xx

xx

•q m

B

v

E

•FEFB

FE = q E and FB = q v B

If FE = FB the particle will movefollowing a straight line trajectory

q E = q v B

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 ?.

xx xx

xx

•q m

B

v

E

•FEFB

FE = q E and FB = q v B

If FE = FB the particle will movefollowing a straight line trajectory

q E = q v B

v = E / B

So need E pointing to the right.


What if the charged particle has a velocity component along B?

rVzr

B

rVz unchanged

Circular motion inxy plane.

z

x

y

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