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# Lorentz force and velocity filter

In this topic we will discuss about Lorentz force and velocity filter.In which we will know different conditions for Lorentz force and complete explanation of velocity filter .

## Lorentz force-

The force experienced by a charged particle due to both electric and magnetic field at a space called Lorentz’s force .

Suppose a charge particle of mass ‘m’ and charge ’q’ is moving in the both electric and magnetic field of Field intensity ’E’ and ‘B’ respectively.

Then force due to electric field is given as Fe= q E

And force due to magnetic field will be Fb=  q v B sinϴ ( where ϴ is the angle between v and B).

Then, Lorentz force F = FE+FB  = q( E + v B sinϴ)………………Eq.

Special cases –

Case 1- When v , E and B all are collinear

In this case Fe = q E ,

And Fb= o ( since ϴ= 0 ; sin 0 = 0 )

acceleration a = Fe/m =  qE/m ,

case 2 – When v , E and B are mutually perpendicular to each other

then net force F = Fe +Fb = 0

therefore acceleration  a= 0 . It means particle will move without any deflection with the same velocity .

## Velocity filter –

It is an arrangement of cross electric and magnetic field in a space which help to select from a beam , charged particles of given velocity irrespective of their charge and mass.

It consist two slits S1 and S2 placed parallel , with common axis at some distance . Uniform electric field ( E ) and magnetic field(B) applied  which are perpendicular to each other , which is shown in figure. When a beam of charged particles of different charges and masses after passing through S1 enters into the region where crossed electric and magnetic field are present . The particles which is moving with velocity ‘v’ the electrostatic force and magnetic force are equal and opposite , then    q E = q v B   ,

Or, v= E/B

Such kind of particles go without changes its path and filtered out the region through the slits S2. Therefore the particles emerging from S2 will have the same velocity even though their charges and masses may be different.

Velocity filter is used in mass spectrograph which help to find the mass and specific charge of the charged particle.

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# Motion of a charged particle in uniform magnetic field

In this topic Motion of a charged particle in uniform magnetic field we will discuss about how a charged particle moves in uniform magnetic field, when it is projected with some velocity at an angle with magnetic field.

Motion of a charged particle in uniform magnetic field

Suppose a charged particle of mass ‘m’ and charge ‘q’ is projected  with velocity ‘v’ at angle ‘ϴ’ with the magnetic field  ‘B’ as shown in figure (a) . Here v has two components  (v1= vcosϴ ) along x -axis and ( v2= v sinϴ) along y-axis.

Due to v1 it moves along x-axis and due to v2 a force F acts which is given as F = qv2 B = qvBsinϴ. Since this force F acts perpendicular to the velocity and magnitude of v2 not changes hence it moves on a circular path.  As shown in fig(b).

let r is the radius of the circular path , due to these two components of v1 and v2 particle follows a helical path as shown in fig(a) .

As we know when body moves on a circular path the centripetal forc F = mv22/r = mv2sin2ϴ/r ;

Which is provided by magnetic force F= qvBsinϴ

So we can write  mv2sin2ϴ/r = qvBsinϴ;

So, r = mvsinϴ/qB ……………….(1)

or vsinϴ=qBr/m …………………….(2)

angular velocity  ω= vsinϴ/r = Bqr/mr = Bq/m ……………..(3)

here frequency  f=ω/2∏ = Bq/2∏m  ( here frequency is independent of velocity)………….(4)

and time period T = 1/f = 2∏m/Bq ……………………………(5).

The pitch of the helix = vcosϴ x T = v cosϴ 2∏m/Bq .

Special cases –

Case (i) -If ϴ=00 i.e. v1=v and v2=0

So there will be no force and particle will move in the direction of B .

Case (ii) – if ϴ=900 then v1=0 and v2=v  i.e. body will move only in the circular path

And radius of the circular path  r= mv/qB  or  r= v/(q/m)B .

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# Newton’s second law of motion is the real law of motion

Before to know Newton’s second law of motion is the real law of motion we must know all three equations of motion.

Second law of motion  is the real law of motion

The question may be asked as , show that second law of motion is the real law of motion .

***This question is very important for the class 11th examination point of view .

To show it we must prove two things (i) First law contained in the second law and (ii) Third law contained in second law.

(i) ) First law contained in the second law –  As we know according to Newtons second law of motion

Force F = Ma ,

If there is no external force

then F = 0

Or, Ma =0    then we can say    M=0 or a=o , but mass of the body can’t be zero.

Therefore only possible thing is that only acceleration is zero .

But , we know acceleration a= v-u/t ( where , v- final velocity and  u- initial velocity) .

So we can write , v-u/t = 0  i.e v-u=0 or v=u

Since V (final velocity) = u ( initial velocity).

So we can say if body initially is in rest will be in rest and if it is moving with some velocity it will remain in the motion with the same velocity( which is the statement of first law ) . So we can say first law contained in the second law .

To watch the video of this topic Newton’s second law is the real law of motion , click on the link given below-

(ii) Third law contained in the second law –  suppose an isolated system of two objects A and B , they interact mutually with one another . Then obviously they will apply force on each other .

Let,  FAB – Force on A due to B ,

And  FBA – Force on B due to A .

Let dpa and dpb is the change in momentum in A and B respectively in small time dt .

Then from Newton’s second law –

FAB = dpA/dt   ………………………(i)

and, FBA= dpB/dt …………………..(ii)

FAB+ FBA= dpB/dt + dpA/dt   ……….(iii)

In the absence of external force , the rate in change in momentum of the system will be zero .

i.e. dpB/dt + dpA/dt  = 0

Therefore , FAB+ FBA=0 ;

Or, FAB= -FBA ( which is third law )

So we can say Newtons third law contained in the second law .

Therefore we can say Newton’s second law of of motion is the real law of motion.

What is the syllabus removed in class 12th physics (CBSE) for 2021 board examination , click on the link given below-

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# Apparent weight of a man in a lift

In this topic we will discuss about , a man of mass m standing in a lift . What will be its apparent weight of a man in a lift when lift has some motion with uniform acceleration or velocity?

Apparent weight of a man in a lift  (elevator)  –

Suppose a man of mass ‘m’ is standing on a weighing machine placed inside a lift . initially when the lift is in rest weight of the man acts vertically downward direction which acts on the   weighing machine and machine offers a resistance ‘R’ against its weight shows the actual weight which is mg .

Case i – When the lift is in rest or moving with uniform velocity (fig-c)- In that case there is no acceleration in the lift and man , and hence reaction force is equal to the weight of the man.

In that case , R  = mg

i.e  Apparent weight = Actual weight

Case ii – When lift moves upward with acceleration ‘a’ fig (a).

Since there is upward motion then  ;

So  R-mg =ma

Or , R = mg+ma = m(g+a)

i.e. R ( apparent weight) > mg(actual weight) ;

i.e  Weighing machine shows more weight the actual .

Case iii – When lift moves downward with acceleration ‘a’ fig (b).

Since there is downward motion then  ;

So  mg-R =ma

Or , R = mg-ma = m(g-a)

i.e. R ( Apparent weight) < mg(Actual weight) ;

i.e  Weighing machine shows less weight the actual .

case iv– When the lift falls freely –  In that case lift accelerates downward with acceleration ‘a=g’

So we can write , R= mg-ma = mg-mg ( since a=g) = o

Show apparent weight of the man will be zero . So person feel’s weightlessness.

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# Horse and cart problem

This is the topic  Horse and cart problem which explain the one of the important illustration of Newton’s third law of motion.

But before to learn the topic Horse and cart problem students must know, Newton’s third law of motion. To get the notes on Newton’s third law of motion click here-

To get the Notes on Newton’s first law and inertia click here-

To get the Notes on Newton’s second law and impulse click here-

Horse and cart problems –

Suppose a cart is connected  with a  horse by a light inextensible string. When horse start to walk horse press the ground F ( action ) with their feet in the slightly backward direction and ground  exert an equal reaction R , as shown in figure . then R has two component H = R cosϴ in horizontal direction  and V= R sinϴ in vertical direction . V balance the weight of the horse and H makes the horse to move in forward direction . While pulling the cart produces a tension T in the string .

If H > T then cart starts moving in the forward direction with acceleration ‘a’ .

Net force acting on the horse = H-T= ma.( where m is the mass of the horse)……………..(i)

If f is the frictional force between tyre of cart and ground then for cart we can write

T-F= Ma  ( M is the mass of the cart ) …………………………(ii)

From equations (i) and (ii) we get ,

H-f = ( M+m ) a ;

So, acceleration   a  =  H-f/ (M+m ) ;

The whole system will move if H>f ;

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# Newton’s third law of motion

In this topic we will discuss about Newton’s third law of motion and some important illustrations of Newton’s third law of motion .

Before to go with this topic students must know Newton’s first and second law .

Newton’s third law of motion-

According to this law every action has equal and opposite reaction.

Forces in nature always occur between pairs of bodies. Let two objects A and B colliding as shown in figure.

FAB force on A due to B ,and FBA force on B due to A .

Then according to Newton’s third law of motion FAB = -FBA .

For example- When a man walk on the ground then man push the road backward and road push the man in forward direction. This example shows , single force never exist and force always exist in pair.

Important point of Newton’s third law we must know

(i) Action and reaction always acts on different bodies.

(ii) No action occur without reaction.

(iii) Action and reaction can’t cancel each other.

(iv) action and reaction may be mechanical, gravitational, electrical or of any other nature.

Some important illustration of Newton’s third law of motion.-

1. Object kept on the table – suppose an object of mass ‘m’ placed on the table , then the weight ‘W’ applying a force equal ti its weight W = mg,  on the surface of table , and table applying a reaction R on the book , In this condition W = -R.

1. 2. Walking of a man on the ground –

While walking we press the ground F ( action ) with our feet in the slightly backward direction and ground  exert an equal reaction R , as shown in figure . then R has two component H = R cosϴ in horizontal direction  and V= R sinϴ in vertical direction . V balance the weight of the man and H makes the man to move in forward direction .

1. It is difficult to walk on the slippery ground or sand-  It is because we are unable to push such a ground sufficiently hard . As a result force of reaction is not sufficient to help a person move in forward direction.
2. While swimming , a person pushes water with his hand in the backward direction and water in turns reacts and pushes the swimmer in forward direction.
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# Newton’s second law of motion and Impulse

In this topic Newton’s second law of motion and Impulse  , we will explain Newton’s second law of motion , expression for force, Impulse and application of concept of impulse . The reasoning questions of this topic and numericals ( specially based on Impulse) is very important for class examination and also for other competitive examination like JEE/NEET .

But before to know this topic, it is necessary the students must  have the concept of Newtons first law of motion and inertia .                                                                                                    For  the notes on Newton’s first law and inertia click here-

### Newtons second law of motion and impulse –

As we know from Newton’s first law of motion ‘every body remain in its position of rest or of uniform motion in a straight line , till no external force acts on it’ . Just think what will happen when external force acts on the body . This effect of  acting a force on the body is described by Newton’s second law of motion .

To watch the video on Newton’s second law – click on the link given below-

Newtons second law of motion ( Force law)-

Newton’s second law of motion states that , when an external force applied on a body , then rate in change in momentum is directly proportional to the applied force , and change takes place in the direction of applied force . this law is also known as the law of force.

We have to take care of its two parts (i) rate of change of momentum is directly proportional to the applied force and , (ii) change in momentum occurs in the direction of applied force .

Expression of force from newtons second law of motion –

Let a body of mass ‘m’ moving with velocity ‘v’ .

Then , its linear momentum will be  p = mv ;

As newton’s second law states  force F α dp/dt   ;  or  F= k dp/dt  ……………..(i)

But here ,  p=mv .

So putting this value of p in equation (i) we get

F=k d(mv)/dt =k m (dv/dt) = k ma; [where dv/dt=a( acceleration) ],

Experimentally it is found k=1

So we can say,  F =ma ;

Unit of force is  kg-m/s2  or kg-ms-2 OR Newton (N)

Its dimension is  [ M L T-2] ;

IMPULSE-

To know the impulse we must know . What is impulsive force?

Impulsive force – It a large force acts on a body for short time which is the cause of change in momentum . example-  Blow a hammer on a nail, force exerted by a bat when it hits a ball, etc.

Impulse – When a large force acts on a body for short time which is the cause of change in momentum, then the product of force and time for which it acts on the body is called impulse or Impulse of force which is equal to the change in linear momentum of the body.

So we can write impulse I = Fav.  = Pf – Pi = Δp

Proof–   From Newton’s second law we know ;

Force F = dp /dt  ;

But momentum P = mv ( m- mass of the body and v is its velocity);

So we can write  F dt = dp ;

Integrating both sides with appropriate limits we get ,

∫ Fdt =∫dp = p2-p1

Or ,  Fav.t = p2-p1  = I

So we can say if the force varying with time ( or force is the function of time ) then we can write

Impulse I = ∫F dt .

If we plot the force v/s time curve , then the area under the curve gives the impulse .

A. When a constant force acts on a body then force v/s time graph is given below . the area under the line AB gives the impulse .

B. When a variable force acts on a body then force v/s time graph is given below . the area under the line ABC gives the impulse ;

The unit and dimension of the impulse is same as linear momentum .

Application of concept of impulse – ( It also work as reasoning questions in your class examinations)

(a) Automobiles are provided with shockers.

(b) China wares are packed under straw paper.

(c) A cricket players lower his hand while catching a ball.

(d) A person falling from a certain height gets more injuries when he falls on a cemented or tough floor than when he falls from a heap of sand.

If we goes to the reason of all the points mentioned above then answers of all the questions will be same .

As we know,  impulse = F x t = change in momentum

For the same impulse force F = impulse / time . i.e. F α 1/t

when time of impact increases due to some processes then force acts on the body decreases .

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# Units and measurement

In this topic we will discuss about, type of units, system of units ,  name and define of all fundamental units . Also the abbreviations in power of tens. one of the important thing given is chart of physicist and their discoveries.

Units and measurement

Physical quantity-

All those quantities which can be measured in terms of which the laws of physics can be expressed are called physical quantity. Example- mass , speed, force , power etc.

Physical quantities are of two types- fundamental and derived

1. Fundamental quantities

The physical quantities which is independent of other physical quantities and are not usually defined in terms of other physical quantities are called fundamental quantities. These are  , mass, length, time, electric current, temperature ,luminous intensity and amount of substance.

1. Derived quantities-

The physical quantities which can obtained using other physical quantities are called derived units. Example – velocity, force , power etc. or we can say all the physical quantities other than seven fundamental quantities are derived units.

UNITS-

The standard amount of a physical quantity chosen to measure the physical quantity of the same kind are called physical unit.

To represent a physical quantity we need numeric value and its units .i.e.

Physical quantity  Q = n U ( where n represent numeric value and U is the unit) . So we can define unit is the thing used to identify and measure a physical quantity .

There are two types of unit .Fundamental and derived units .

Fundamental units

The physical unit which neither be derived from one another and it can not be resolved into more simpler units called fundamental units. There are seven fundamental units which are the units of These are  , mass, length, time, electric current, temperature ,luminous intensity and amount of substance .

Derived units –  All the units can be expressed using fundamental units is called derived units .

Example unit of speed = distance/time =m/s …etc .

System of units – A complete set of units which is used to measure all kind of fundamental and derived quantities is called system of units.

(i) cgs system- It is based on centimetre, gram and second as the fundamental unit of length, mass and time respectively.

(ii) MKS system – – It is based on metre, kilogram and second as the fundamental unit of length, mass and time respectively.

(iii) FPS system – – It is based on foot, pound and second as the fundamental unit of length, mass and time respectively.

(iv)SI ( the international system of units) system .

Definition of basics units –

1. Meter (m) – It is the SI unit of length, One meter is defined as the path travelled by light in vacuum in 1/ 299,792,458 seconds ,
2. kilogram (kg) – It is the SI unit of mass .- It is the mass of prototype cylinder of platinum-iridium alloy .
3. Second(s)- It is the SI unit of time . On e second is the duration of 9,192,631,770 period of the radiation between two levels of the ground state of the Cesium-133 atom.
4. Ampere (A) – It is the unit of electric current. It is the force 2 x 10-7 Newton between two parallel current carrying wire placed 1m away is of unit length . To know more on Ampere click here-
5. Kelvin (K) – It is the SI unit of temperature. One kelvin is the fraction 1/273 of the thermodynamic temperature of the triple point of water.
6. Candela(cd)- It is the SI unit of luminous intensity. It is the intensity of a source that emits monochromatic radiation of frequency 540 x 1012 Hz , and that has the radiant intensity 1/683 watt per steradian in that direction.
7. Mole (mol)- It is the unit of amount of substance . .It is the amount of substance which contain as many elementary entities as there are atom in 0.012 kg of C-12 isotopes.

Supplementary SI units –

(a). Radian (rad)-  It is the plane angle subtended at the centre of a circle by an arc equal in length to the radius of the circle . ϴ= arc/radius .

(b). Steradian (sr)- It is defined as the solid angle subtended at the centre of the sphere by a surface of the sphere equal in area to that of a square , having each side equal to the radius of the sphere . sr= surface area / radius2 .

Some important thing ( multiple , prefix and symbol) in the power of ten.

Some great physicist and their discoveries ( This topic is important for objective point of view)-

Next topic after this topic units and measurement students have to learn DIMENSIONS AND USES OF DIMENSIONS

Categories

# Solenoid and Toroid

Before to know Solenoid and Toroid  , students must know Ampere’s circuital law , proof of Ampere’s circuital law and its application.        To know all these thing click here-

In this topic we will discuss about  one of the application of Ampere’s circuital law . We will define solenoid and Toroid , we will find the magnetic field due to Solenoid and Toroid .

Solenoid

It is the closely wound coil in the form of helix . its length is very large as compared to its diameter.

Magnetic field due to a solenoid

Let current I is flowing through the coil , each turn of solenoid regarded as a circular loop carrying current which produces a magnetic field . Total magnetic field is vector sum of magnetic field due to current through all the turns in the coil .

Let n be the number of turns per unit length of the solenoid . Consider a rectangular loop PQRS  near the middle of the solenoid as shown in figure.

PQ=L . hence total numbers of turn in length L = nL .

The line integral of magnetic field over the closed path PQRS is ,

At a point near the end of the solenoid magnetic field B = μ0nI/2 .

If the solenoid is filled by material of permeability μ in side then magnetic field B = μnI = μ NI/L.

If we draw a plot magnetic field B vs r (distance) from the centre of the solenoid we get the following curve.

Toroid –

Toroid is the endless solenoid in the form of ring . or we can define ‘The toroid is the hollow circular ring on which a large number of insulated turns of a metallic wire are closely wound’. As shown in figure below.

Magnetic field due to current in a toroid–  Let n be the number of turns per unit length of the toroid , I be the current flowing through the toroid . When current passes through the solenoid magnetic field of constant magnitude setup in side the turn of toroid in the form of circular magnetic field . We draw three circle having radii r1,r2 and r3 as shown in fig (b). Let B1 is the magnetic field along loop 1  then using Ampere’s law –

The magnetic field at any point inside the empty space surrounded by toroid or outside the toroid magnetic field is zero .

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# Ampere’s circuital law

In this topic we will discuss about Ampere’s circuital law and proof of Ampere’s circuital law ( using Biot- Savart’s law).We will also discuss the  applications of  Ampere’s circuital law (Magnetic field due to infinite long straight wire carrying current,magnetic field due to current through very long circular cylinder or thick wire, solenoid and Toroid)

Ampere’s circuital law –

According to this law the line integral of the magnetic field around any closed path in free space is equal to μ0 times the total current passing through the surface enclosed by the closed path .

To watch the video of Ampere’s circuital law click on the link given below-

Applications of Ampere circuital law –