pak physics notes.......

Resistance and resistivity ( By pak physics)

2011-11-18T21:16:00.000-08:00

Opposition offered by the material of conductor in the flow of electric current is called resistance.
Resistance opposes the flow of current through a conductor. Resistance of a conductor is due to the collision of free electrons with the atoms of the conductor.
It is denoted by “R”

(1) Length of conductor

Resistance of a conductor is directly proportional to the length of conductor .

R a L…………….(a)

(2) Area of cross section of conductor

Resistance of a conductor is inversely proportional to area of cross section of conductor.

R a 1 / A……………….(b)

Combining (a) and (b)

R α L/A

R = rL/A

	Where r = resistivity of material of conductor
	RESISTIVITY
	Resistivity is an electrical property of material . It is defined as the resistance of a material or conductor of 1 cubic meter volume. Or It is the resistance of a conductor of unit length and unit area. Or Resistivity of a conductor is the resistance of 1 meter long conductor whose area of cross section is I meter square
	Unit: R = rL/A r = ohm x m Different materials have different values of resistivity. A very high value of resistivity indicates high electrical resistance

Electric Current (By pak physics notes)

2011-11-18T21:00:00.000-08:00

"Electric current is defined as the amount of electric charge passing through a cross section of a conductor in unit time."

"The rate of flow of electric charge through a cross section of a conductor is called Electric Current".

Mathematically

Electric current = electric charge / time

I = Q/t

Unit of electric current is AMPERE. 1 ampere = 1 coulomb / 1 sec

AMPERE
In S.I system unit of electric current is ampere. Ampere is defined as: “Current through a conductor will be 1 ampere if one coulomb of electric charge passes through any cross section of conductor in 1 second.” 1 ampere = 1 coulomb / 1 sec

There are two types of current.
ELECTRONIC CURRENT
Electronic current flows from negative to positive terminal.
CONVENTIONAL CURRENT
Direction of conventional current is taken from higher potential to the lower potential. (www.pak-physicsnotes.blogspot.com)

Capacitance of capacitor (By pak physics notes)

2011-11-18T20:52:00.000-08:00

The ability of a capacitor to store electric charge between its plates is its capacity or capacitance.
DEFINITION
The capacitance of a capacitor is defined as: "The ratio of electric charge stored on any one of the plates of capacitor to potential difference between the plates."

Mathematically
C = Q / V
UNIT OF CAPACITANCE
In S.I. system unit of capacitance is Coulomb / volt OR Farad Farad is a large unit therefore in general practice we use small units (1) uF (microfarad) 1 uF = 1x10^-6 F (2) uuF (Pico farad) 1 uuF = 1x10^-12^{F For latest information , free computer courses and high impact notes visit :}^{www.citycollegiate.com}
CAPACITANCE OF A PARALLEL PLATE CAPACITOR
Consider a parallel plate capacitor as shown below:
Let The area of each plate = A The separation between plates = d Medium = air Surface density of charge on each plate = s The electric field intensity between the plates of capacitor is given by
E = s / eo
Potential difference between the plates of capacitor can be calculated by the following relation
V = Ed Putting the value of "E" V = (s/eo) x d But s = Q/A V = Qd/A.eo AeoV = Qd…………(a)
We know that The electric charge stored on any one of plate of capacitor is
Q = CV……………(b) Putting the value of Q in (a) AeoV = CVd Cd = Aeo
C = Aeo/d (www.pak-physicsnotes.blogspot.com)

Columb's Law (By pak physics notes)

2011-11-18T20:49:00.000-08:00

STATEMENT

The electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of charges inversely proportional to the square of distance between them.

Explanation

Consider two point charges q1 and q2 at a placed distance of r from each other. Let the electrostatic force between them is F.

According to the first part of the law:

F α q₁q₂

According to the second part of the law:

F α 1/r²

Combining above statements:

F α q₁q₂/r²

F =k q₁q₂/r^{2 --------------------(1)}

Where k is the constant of proportionality.

Value of K is equal to 1/4pe₀

where eo is permittivity of free space .Its volume is 8.85 x 10^-12 c²/Nm².

Thus in S.I. system numerical value of K is 8.98755 x 10⁹ Nm²c^-2.

Putting the value of K = 1/4pe₀in equation (i)

F = 1/4pe₀q₁q₂/r²

FORCE IN THE PRESENCE OF

DIELECTRICMEDIUM

If the space between the charges is filled with a non conducting medium or an insulator called "dielectric", it is found that the dielectric reduces the electrostatic force as compared to free space by a factor (e_r) called DIELECTRIC CONSTANT. It is denoted bye_r . This factor is also known as RELATIVE PERMITTIVITY. It has different values for different dielectric materials.

(www.pak-physicsnotes.blogspot.com)

Gauss's Law (By pak physics notes)

2011-11-18T20:24:00.000-08:00

Gauss’s law is a quantitative relation which applies to any closed hypothetical surface called Gaussian surface to determine the total flux (Ø) through the surface and the net charge(q) enclosed by the surface.

Statement

"The total electric flux through a closed surface is equal to 1/e_o

times the total charge enclosed by the surface."

PROOF

Consider a Gaussian surface as shown below which encloses a number of point charges q₁,q₂,q₃…….q_n
Draw imaginary spheres around each charge. Now we make use of the fact that the electric flux through a sphere is q/e_o.

Flux due to q₁ will be Ø₁ = q₁/e_o

Flux due to q₂ will be Ø₂ = q₂/e_o

Flux due to q₃ will be Ø₃ = q₃/e_oFlux due to q_n will be Ø_n = q_n/e_o

	Hence the total flux Øe will be the sum of all flux i.e
	Ø = Ø₁ + Ø₂+ Ø_{3 +}Ø₄……….+ Ø_n
	Ø = q₁/e_o+ q₂/e_o+ q₃/e_o+……….. +q_n/e
	Ø = 1/e_o(q₁+ q₂+ q₃+……….. +q_n)
	Ø = 1 /e_o x ( total charge)
	This shows that the total electric flux through a closed surface regardless of its shape or size is numerically equal to 1 /e_otimes the total charge enclosed by the surface. (www.pak-physicsnotes.blogspot.com)

Elstic and inelastic collision (pak physics notes)

2011-11-18T08:04:00.000-08:00

Elastic collision
An elastic collision is that in which the momentum of the system as well as kinetic energy of the system before and after collision is conserved.
Inelastic collision
An inelastic collision is that in which the momentum of the system before and after collision is conserved but the kinetic energy before and after collision is not conserved.
Elastic collision in one diamension

Consider two non-rotating spheres of mass m₁ and m₂ moving initially along the line joining their centers with velocities u₁and u₂in the same direction. Let u₁is greater than u₂. They collide with one another and after having an elastic collision start moving with velocities v₁and v₂ in the same directions on the same line.

Momentum of the system before collision = m₁u₁ + m₂u₂

Momentum of the system after collision = m₁v₁ + m₂v₂

According to the law of conservation of momentum:

m₁u₁ + m₂u₂= m₁v₁ + m₂v₂
m₁v₁ – m₁u₁= m₂u₂ – m₂v₂
m₁(v₁ – u₁) = m₂(u₂ – v₂) -------(1)

Similarly

K.E of the system before collision = ½ m₁u₁² + ½ m₂u₂²
K.E of the system after collision = ½ m₁v₁² + ½ m₂v₂²

Since the collision is elastic, so the K.E of the system before and after collision is conserved .

Thus

½ m₁v₁² + ½ m₂v₂² = ½ m₁u₁² + ½ m₂u₂²
½ (m₁v₁² + m₂v₂²) = ½ (m₁u₁² + ½ m₂u₂²
m₁v₁²-m₁u₁²=m₂u₂²-m₂v₂²
m₁(v₁²-u₁²) = m₂(u₂²-v₂²)
m₁(v₁+u₁) (v₁-u₁) = m₂(u₂+v₂) (u₂-v₂) ------- (2)

Dividing equation (2) by equation (1)

V₁+U₁ = U₂+V₂

From the above equation

V₁=U₂+V₂-U₁_________(a)

V₂=V₁+U₁-U₂_________(b)

Putting the value of V₂ in equation (1)

m₁ (v₁-u₁) =m₂ (u₂-v₂)
m₁ (v₁-u₁) =m₂{u2-(v₁+u₁-u₂)}
m₁(v₁-u₁)=m₂{u₂-v₁-u₁+u₂}
m₁(v₁-u₁)=m₂{2u₂-v₁-u₁}
m₁v₁-m₁u₁=2m₂u₂-m₂v₁-m₂u₁
m₁v₁+m₂v₁=m₁u₁-m₂u₁+2m₂u₂
v₁(m₁+m₂)=(m₁-m₂)u₁-2m₂u₂

In order to obtain V₂ putting the value of V₁ from equation (a) in equation (i)

m₁(v₁-u₁) = m₂(u₂-v₂)

m₁(u₂+v₂-u₁-u₁)=m₂(u₂-v₂)
m₁(u₂+v₂-2u₁)=m₂(u₂-v₂)
m₁u₂+m₁v₂-2m₁u₁=m₂u₂-m₂v₂
m₁v₂+m₂v₂=2m₁u₁+m₂u₂-m₁u₂
v₂(m₁+m₂)=2m₁u₁+(m₂-m₁)u₂

(www.pak-physicsnotes.blogspot.com)

Young double slits experiment (By pak physics notes)

2011-11-18T08:00:00.000-08:00

The first practical demonstration of optical interference was provided by THOMAS YOUNG in 1801. His experiment gave a very strong support to the wave theory of light.

Arrangement of Experiment

'S' is a slit, which receives light from a source of monochromatic light. As 'S' is a narrow slit so it diffracts the light and it falls on slits A and B. After passing through the two slits, interference between two waves takes place on the screen. The slits A and B act as two coherent sources of light. Due to interference of waves alternate bright and dark fringes are obtained on the screen.

Experiment

Let the wave length of light = l
   Distance between slits A and B = d
   Distance between slits and screen = L
   Consider a point 'P' on the screen where the light waves coming from slits A and B interfere such that    PC=y. The wave coming from A covers a distance AP=r1 and the wave coming from B covers a distance    BP=r2 such that PB is greater than PA.

Path difference = BP-AP = BD

S = r2-r1 = BD

In right angled DBAD

Sinq = BD/AB
Or
sinq = s/d
Or
S = dsinq -------(1)

Since the value of 'd' is very very small as compared to L, therefore, q will also be very small. In this condition we can assume that :

Sinq = tanq
From (1)
S = dtanq ---(2)
In right angled DPEC
Tanq = PC/EC = y/L
Putting the value of tanq in eq. (2), w get
S = dy/L
Or

y = SL/d -----(3)

FOR BRIGHT FRINGE
For bright fringe S = ml -----(3) Therefore, the position of bright fringe is:
_{y = m}_l_L/d

FOR DARK FRINGE AT P
For destructive interference, path difference between two waves is (m+1/2)l ----(3) Therefore, the position of dark fringe is: y = (m+1/2)lL/d
FRINGE SPACING
The distance between any two consecutive bright fringes or two consecutive dark fringes is called fringe spacing. Fringe spacing or thickness of a dark fringe or a bright fringe is equal. It is denoted by Dx.
Consider bright fringe. y = mlL/d For bright fringe m=1 y1 = (1)lL/d for next order bright fringe m=2 y2 = (2) lL/d fringe spacing = y2 - y1 or Dx = (2)lL/d - (1)lL/d Dx =lL/d (2-1) Dx = lL/d Similar result can be obtained for dark fringe. (www.pak-physicsnotes.blogspot.com)

Force on a current carrying conductor moving in uniform magnatic field (By pak physics notes)

2011-11-18T07:44:00.000-08:00

Consider a conductor of length "L", area of cross section "a" placed in a uniform magnetic field of "B".
Let there are "n" number of charges (e) per unit volume and an electric current "I" is passing through it.

	Volume of conductor = V = area x length = a x L Number of electrons = N = n . Volume Charge on each electron = e Total charge = q= Ne We know that the force experienced by a charge in a uniform magnetic field is

	F = Q (V x B )
	Putting the value of "q" F = Ne (vBSinq) F = nV. e(vBSinq) F = ne . aL (vBSinq) F = naLe (v B Sinq)
	Now consider the length of conductor (L) as a vector quantity in the direction of velocity vector.
	F = nave(Lv B Sinq) F = nave(LB Sinq)
	Since S = vt v = s/t v = L/t
	Putting the value of v
	F = na(L/t)e (L B Sinq) F = naLe/t (L B Sinq)
	Since naLe = q Therefore,
	F = q/t (L B Sinq)
	Also q/t = I
	F = I (L B Sinq)
	F = ILB Sinq
	The expression clearly indicates that the force acting on the wire when placed in a uniform magnetic field is perpendicular to the plane formed by the L and B. i.e. F is perpendicular to L and also perpendicular to B.

(www.pak-physicsnotes.blogspot.com)

Torque on Current carrying coil In magnatic field (pak physics notes)

2011-11-18T07:35:00.000-08:00

Consider a rectangular coil of wire through which an electric current 'I' is being passed. The coil is placed in a uniform magnetic field of magnitude "B".

	The vertical side AB of the coil experiences a force F which is directed perpendicular into the paper. There is an equal and opposite force F on side CD. Since the plane of coil is parallel to the magnetic field therefore there is no force on either side AC and BD. Force on AB and CD will produce torque in these arms.


	In this case L perpendicular to B i.e. q = 90, hence
	F = BIL (1) F = BIL _AB = b/2 (BIL)…………(1) Similarly _CD = b/2 (BIL)…………(2)
	These two torque result in a couple in the coil and the coil starts moving (rotating) in magnetic field. Since both the torque are trying to rotate the coil in the same direction, therefore the total torque produced in the coil will be the sum of all the individuals torque.
	= _AB + _CD = b/2 (BIL) + b/2 (BIL) = 2/2 b (BIL) = IB (Lb)
	but Lb = area of coil = A
	= IBA
	If there are "N" number of turns of wire in the coil then the torque will becomes N times
	= BINA
	When the coil makes an angle a with the direction of magnetic field, the torque will become
	= BINA Cosa

(www.pak-physicsnotes.blogspot.com)

Kinetic Energy (By pak physics notes)

2011-11-18T06:39:00.000-08:00

Kinetic Energy
'' The Energy of a body Due to its motion is known as Kinetic Energy''.
FORMULA
K.E. = 1/2 mv²
Kinetic energy depends upon the mass and velocity of body.If velocity is zero than K.E. of body will also be zero.Kinetic Energy is a scaler quantitiy.
Proof
Consider a body of mass "m" starts moving from rest. After a time interval "t" its velocity becomes V.

If initial velocity of the body is Vi = 0 ,final velocity Vf = V and the displacement of body is "d". Then

First of all we will find the acceleration of body.

Using equation of motion

2aS = V_f² – V_i²
Putting the above mentioned values
2ad = V²– 0
a = V²/2d

Now force is given by
F = ma
Putting the value of accleration
F = m(V²/2d)
As we know that
Work done = Fd

Putting the value of F
Work done = (mv²/2d)(d)
Work done = mV²/2
OR
Work done = ½ mV²

Since the work done is motion is called "Kinetic Energy"
i.e.
K.E. = Work done
OR
K.E. =1/2mV².

(www.pak-physicsnotes.blogspot.com)

Law of conversation of energy (By pak physics notes)

2011-11-18T06:27:00.000-08:00

Law of conversation of energy
This law state the
'' Energy can neither be destroyed nor be created but can change from one form to anotheer''
Proof by simple pandulam

Energy at point ''A''

At point ‘A’ velocity of the bob of simple pendulum is zero. Therefore, K.E. at point ‘A’ = 0. Since the bob is at a height (h), Therefore, P.E. of the bob will be maximum. i.e.
P.E. = mgh.
Energy total = K.E. + P.E
Energy total = 0 + mgh
Energy total = mgh This shows that at point A total energy is potential energy.

Energy at point ''M''

If we release the bob of pendulum from point ‘A’, velocity of bob gradually increases, but the height of bob will decreases from point to the point. At point ‘M’ velocity will become maximum and the height will be nearly equal to zero.
Thus ,
K.E. = maximum = 1/2mV² but P.E. = 0.
Energy total = K.E. + P.E
Energy total = 1/2mV² + 0
Energy total = 1/2mV²
This shows that the P.E. at point is completely converted into K.E. at point ‘M’.

Energy at point ''B''

At point M the bob of Pendulum will not stop but due to inertia, the bob will moves toward the point ‘B’. As the bob moves from ‘M’ to ‘B’, its velocity gradually decreases but the height increases. At point ‘B’ velocity of the bob will become zero.
Thus K.E. at point ‘B’ = 0 but P.E. = max.
P.E. = mgh.
Energy total = K.E. + P.E.
Energy total = 0 + mgh
Energy total = mgh
This shows that at point B total energy is again potential energy.

(www.pak-physicsnotes.blogspot.com)

Density of Earth (By pak physics notes)

2011-11-18T06:16:00.000-08:00

We know that the earth is composed of numerous substances such as metals, non-metals, water and other materials. Therefore we can not determine exact density of the earth .For our convenience, here, we are assuming that the earth is made of homogeneous substance and it has a perfect spherical shape, however which is not true.

We know that the density of a substance is given by:

density = mass / volume

For earth:

r = M_e / V_e

Let the earth is a sphere of homogeneous composition , therefore its volume is given by:

V_e = 4/3(pR³_e)

	Hence the average density of the earth is:

Mass of Earth (by pak physics notes)

2011-11-18T06:05:00.000-08:00

Consider a body of mass ‘m’ placed on the surface of the earth. Let the mass of the earth is ‘M_e’ and radius of the earth is ‘R_e’ .Neglecting the radius of body if compared with that of the earth.

Gravitational force of attraction between earth and body is

F = G m M_e/ R_e²^........(1)

We know that the force of attraction of the earth on a body is equal to weight the weight of body. i.e

F = W ........................(2)

thereforeComparing (1) and (2), we get,

W = G m M_e/ R_e²

But W = mg

mg = G m M_e/ R_e²

or
g = G M_e/R_e²

or
M_e = g x R_e²/G

From astronomical data:

g= 9.8 m/s²
R_e= 6.38 x 10⁶ m
G = 6.67 x 10^-11 N-m²/kg²
Putting these values in the above equation.

M_e = 9.8 (6.38 x 10⁶)²/6.67 x 10^-11
or

M_e = 6x10²⁴ Kg

(www.pak-physicsnotes.blogspot.com)

Centripital ccelration

2011-11-18T05:53:00.000-08:00

Centripital force

The force which keeps the body to move in a circular path is called centripital force.

The force which bend the normally straight path in to circular path in known as centripital force.

It is denoted by F_c . its formula is

F_c = mv²/r

Centripital accelration

If an object is moving in a circle with constant speed, then direction of its motion is continuously changing. Due to change in direction of body an acceleration is produced in the object The directionof acceleration is towards the center of circle. This acceleration is called is called "centripetal acceleration".

It is denoted by a_c.

Expression For F_c

Suppose a body is moving in a circle of radius r with constant speed V. Let it travels from point A to B
in a time interval of Dt sec and covers a distance DS meter. Let velocity vector of the body at point
A be V₁ and at point B be V₂.Angle traced at the center during this interval Dt sec is say Dq .
Take a point O outside circle, draw velocity vector OP and OQ equal and parallel to V₁ and V₂. Join PQ from vector addition head and tail rule OQ i.e. V₂ is resultant of OP = V₁ and PQ = DV. It means PQ is the change in velocity DV during interval Dt sec.

DV = V₂ – V₁

Angle POQ is equal to angle at the center of circle becomes V₁ and V₂ are perpendicular to radius lines at A and B. From geometry DACB and DOPQ are similar because both are in closed triangles and Dq are same.

By similer properties

D V / V = D S / r

D V = V. D S/r

Dividing both sides by D t

D V / D t = V/r . D S/D t

When Dt is very small D tè 0

D V / D t is lim DV / Dt

D tè 0

V / r lim DS / Dt

D tè 0

a_c = V / r * V

a_c = V² / r

This is the magnitude of centripital accelration.

By 2nd law of motion

F_c = ma

F_c = m(v²/r)

F_c = mv²/r

In angular motion

V =rω

Therefore

F_c = mr²ω²/r

F_c= mrω²

(www.pak-physicsnotes.blogspot.com)

Motional EMF (pak-physicsnotes.blogspot.com)

2011-11-17T23:04:00.000-08:00

Motional Emf

The emf induced by the motion of conductor across the magnatic field is called motional emf.

Explanation

Consider a rod of length ‘’L’’ placed between two matel rails saprated by the distance of L. A galvanometer is connected between c and d. This forms a complete abcd as show in fig.

When rod is pulled towards right with velocity v, the galvanometer shows that current is flowing through loop. The rod acts as a source of emf.

E = V_b –V_a= ΔV

We know that force experience by the charge moving perpandicular to the magnatic field is given by

F = q(V x B)

F= qVBSinѲ

F = qVB (Ѳ = 90)

According to right hand rule the direction of force is from a to b. The current is flowing anticlock wise. After some times the concentration of charges at the top increase and deficiency occurs at bottom.
The electric field will be set up from top to bottom. The electric field exert a force on the charge.

At equilibrium

F_e = F_b

qE = qVB

E = VB ____ (1)

As we know that electric field intensity is equal to the negative of potential gradient . So

E = - Δv/L _____ (2)

Comparing eq 1 & 2

VB = - ΔV/L

VB = -Є/L

Є = - VBL

If Ѳ is the angle between V and B

Є = -VBL SinѲ

(www.pak-physicsnotes.blogspot.com)

Voltmeter (pak-physicsnotes.blogspot.com)

2011-11-17T22:51:00.000-08:00

VOLTMETER

The electrical instrument which is used to measure the potential difference between two points in volts is called voltmeter.Voltmeter is always connected in parallal in the circuit for measuring the potential difference.

CONVERSION

Voltmeter is the modified form of galvanometer. If we connect a high resistance with the galvanometer in series, the galvanometer is coverted into such electrical instrument which is called voltmeter.

Expression

As high resistance R_h is connected in series with the galvanometer, hence voltage in R_h and R_gis different. While same current I_g will flow through them. A small current I_g will required for the metermovement (coil) of galvanometer, So that the full scale deflection is obtain and rest of the current should through the component of the curcuit , Whose potential is required to measure.

Applying OHM’s law

V= IR

V= I_gR_{eq _________________ (1)}

Where

R_eq= R_h + R_g

Put in eq no. (1)

V = I_g(R_h+ R_g)

V = I_gR_h + I_gR_g

I_gR_h= V - I_gR_g

R_h= V - I_gR_g

I_g

(www.pak-physicsnotes.blogspot.com)

Galvanometer (www.pak-physicsnotes.blogspot.com)

2011-11-17T22:33:00.000-08:00

Galvanometer

It is an electrical device used to measure the very small current in milli-ampere.

Principle

When coil is placed in a magnatice field, torque produce which tends to rotate the coil.

Construction

It consist of a coil. The coil is made by wounding the copper wire on a non-magnatic material.The coil is suspended in a u- shaped magnet by a wire. The one end of the coil is attached with the mirror and the second end is attached with helical spring. Both ends of the wire also acts as terminalas shown in fig.

In fig N and S are north and south poles of concave shaped magnet resceptively. D is the iron cylinder. The concave shape and iron cylinder placed inside the coil makes magnatic field radial and strong. The plane of the coil always remains parallal to the magnatic field.

Working

Now when current passed through the coil,torque produces, which deflect the coil. Hence

Deflecting torque = NIABcos α

As α is the angle between plane of the coil and Magnatic field B. This angle may always be zero.

So τ_d = NIAB cos 0

. τ_d = NIAB

Restoring torque

The spring attached with one end of the coil will re-twist the coil towards its orignal position. The restoring torque is directly proportional to the twist angle θ.

Restoring torque θ

. τ_r θ

. τ_r = Cθ

Where C is the constant of proportionality and called twist constant.

(www.pak-physicsnotes.blogspot.com)