Physics Classes

Author: Nayan Jha

Newton’s first law of motion and inertia
Newton’s first law of motion and inertia

In this topic we will discuss about Newton’s first law of motion and inertia . Force , linear momentum and some example of inertia (all type).

Newton’s first law of motion ( law of inertia )-

Before to know the Newton’s first law of motion and inertia we must know –

FORCE– Force is a physical quantity which when applied on a body it actually changes or try to change , shape, size and, position of rest or motion.

INERTIA – It is the property of a body by virtue of which it can’t change itself its state of rest or uniform motion in a straight line . Or we can say ‘inertia is the resistance of change’. this term inertia is first used by Galileo .

Different types of inertia –
1. Inertia of rest – It is the tendency of a body to remain in its position of rest .
Example- A person standing in a bus fall backward when the bus suddenly start moving forward. Dust is removed from a hanging carpet by beating it with a stick, etc.
1. Inertia of motion– It is the tendency of a body to remain in its state of uniform motion in a straight line .
Example-When a running bus stop suddenly the passengers in the bus fall in forward direction. It is dangerous to jump out from a running bus, etc.
1. Inertia of direction– It is the property of the body to restrict the change of direction of motion.
Example- When a bus takes a sharp turn the person sitting in the bus experiences a force acting away from the centre of the curved path .

Linear momentum (momentum) – momentum of a body is the quantity of motion possessed by the body . Mathematically it is equal to the product of mass and velocity of the body .

If ‘m’ is the mass of the body and its velocity is’v’ then it is given momentum P = m v .

It is a vector quantity .

Its unit is kg m/s or Ns.

And its dimension is [M L T^-1] .

Newton’s first law of motion – It is also known as law of inertia. According to 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’ .

This law consist of three parts ;

(i) firs part say a body is in rest continues in the states of rest( inertia of rest).

(ii) second part say that body is in motion continue moves in the same path with same speed (inertia of motion)

(iii)third part say that if a body moving with wit uniform velocity in a st. line then it cant change its direction of motion it self ( inertia of direction ) .

From the above three parts given above explain that how Newton’s first law explain the law of inertia.

Illustration of Newton’s first law of motion –
1. Based on inertia of rest-
(i) When we shake a branch of tree its fruit and dry leaves fall down – It is because when we shake the tree its branches comes in motion but due to inertia of rest fruit or leaves wants to be in rest due to which it separated from the branches of tree and fall down.

(ii) When a horse suddenly start running then the riders on it fall back -It is because initially the rider and horse are in rest , when horse starts running suddenly the part of the rider in contact with the horse comes into the motion but upper part of the body due to inertia in rest remain in rest and the rider falls back.

(iii) A person standing in a bus fall backward when the bus suddenly start moving suddenly forward- -It is because initially the person and bus are in rest , when bus starts moving suddenly the part of the person in contact with the bus comes into the motion but upper part of the body due to inertia in rest remain in rest and the person falls back.

(iv) Dust is removed from a hanging carpet by beating it with a stick- it is because when we hit the carpet it comes in motion but due to inertia of rest dust wants to be in rest due to which it separated from the carpet and fall down.

2.Based on inertia of motion–

(i) When a running bus stop suddenly the passengers in the bus fall in forward direction.- It is because when the bus is in motion then the passenger on it also is in motion , but when bus stop suddenly then the portion of the passenger is in contact with the bus also comes in rest but due to inertia of motion upper part of the passenger remain in motion and hence passengers fall in forward direction or in the direction of the moving bus .

(ii) It is dangerous to jump out from a running bus – .- It is because when the bus is in motion then the passenger on it also is in motion , but when the person jumps out then the lower part of the body comes in contact with the ground and it comes in rest but due to inertia of motion upper portion of the body remain in motion and person fall down in the direction of motion of bus and it becomes the cause of accident.

NEXT TOPIC -Newton’s second law of motion and Impulse. To get the notes on Newton’s second law of motion and Impulse click here-
July 23, 2020
Magnetic field at a point on the axis of a circular coil carrying current

Magnetic field at a point on the axis of a circular coil carrying current

Before to learn about this topic students must know about Biot-Savart law. To learn about this topic click here-

In this topic we will discuss about Magnetic field at a point on the axis of a circular coil carrying current , and using its derivation we can find the magnetic field and we will also discuss about magnetic moment due current carrying coil.

To get the notes on magnetic field at the center of a circular current carrying coil, click here-

Magnetic field at a point on the axis of a circular coil carrying current –

Suppose a circular coil of radius ‘a’ with center ‘O’ . Let current I is flowing through the coil we have to find the magnetic field at point ‘P’ , which is x distance away from the center .

Suppose two small element ‘dl’ of the coil C and D which is diametrically opposite points as shown in figure.

Here PC =PD = √(a²+x²), and we consider <COP = ɸ = <DPO .

As shown in figure dBcosɸ is cancelled by each other , then the net magnetic field dB sinɸ will be in the same side .

Here magnetic field due to small current carrying element dB = (µ₀/4Π) I dl sinθ/r² ; here r=√(a²+x²),

So we can write , dB=(µ₀/4Π) Idl sinθ/(a²+x²) ,

So magnetic field at point p due to the circular loop

Special case 1- when point P lies at the center of the circular coil then , x = 0

Then B= (µ₀/4Π) 2∏nI/a = µ₀nI/2a ,

Case 2 – When point P is far away from the center then a²+x²=x²

Then B= (µ₀/4Π) 2nIA/x³ [ since ∏a² = A (area)]

Here nIA= M (magnetic moment)

So we can write , B= (µ₀/4Π) 2M/x³

So we can define the magnetic moment due to current carrying coil is given as the product of ampere turns and area of current loop . SI unit of magnetic moment is A-m² .

To watch the video related to this topic, Magnetic field at a point on the axis of a circular coil carrying current ( By Nayan jha sir) go to the link given below-

The polarity of magnetic dipole due to the current loop is decided as , if the current from one side is clock wise direction it gives south pole and on another face direction of current is anti-clock wise it gives north pole , as shown in figure-

Case 3- The variation of magnetic field induction with distance of a point on the axis of coil carrying current is given as –

class 12th physics syllabus removed . How it is beneficial for the students , see the video given below-

July 23, 2020
Magnetic field at the center of a circular current carrying coil

Magnetic field at the center of a circular current carrying coil

Before to know about this topic students must know Biot-Savart’s law.

To get the notes on Biot-Savart’s law click here-

Magnetic field at the center of a circular current carrying coil is the one of the application of Biot-Savart’s law . here we will derive the expression for Magnetic field at the center of a circular current carrying coil .

syllabus class 12th physics (2020-2021)

Magnetic field at the center of a circular current carrying coil – Consider a circular coil of radius ‘r’ having center ‘O’. suppose I be the current flowing through the coil , and we have to find the magnetic field at the center .

Suppose a small element ‘dl’ which is the part of coil create a magnetic field dB at the center.

According to Biot-savart’s law dB = (µ₀/4Π) I dl sinθ/r² . but ϴ=90⁰,

So we can write dB = (µ₀/4Π) I dl sin90⁰/r² = dB = (µ₀/4Π) I dl /r² .

Then magnetic field at the center due to complete coil

B=∫ (µ₀/4Π) I dl sinθ/r² ( Taking limit 0 to 2∏)

We get B= dB = (µ₀/4Π) I 2∏r/r² = µ₀ I /2r

For an arc which is making angle ϴ at the center will be given as

B= (µ₀ I /4∏r)(Angle at the center )

Or , B= (µ₀ Iϴ /4∏r) ;

The direction of magnetic field due to current carrying coil may be give by right hand rule , according to it if curled finger shows the direction of current then stretched thumb gives the direction of magnetic field .

To watch the video of related topic click here-

For next topic magnetic field at a point on the axis of a circular current carrying coil . click here-

July 22, 2020
Force on a moving charge in a Magnetic field

Force on a moving charge in a Magnetic field

In this topic Force on a moving charge in a Magnetic field , define the magnetic field and units and dimension of magnetic field .We will also know about Fleming’s left hand rule.

Before to know about this topic Force on a moving charge in a Magnetic field we must learn about Oersted’s experiment and Ampere’s swimming rule. To know about this topic click here –

Force on a moving charge in a Magnetic field –

Suppose a positive charge ‘q’ is moving with velocity ‘v’ at an angle ‘ϴ’ with magnetic field ‘B’. Then experimentally it is found that   force ‘F’ experienced depends on

F α q ………(i)

F α B……………….(ii)

F α v sinϴ …………(iii)

On combining these three equations we get ,

F α q B v sinϴ

Or, F =k q B v sinϴ ; where k is constant of proportionality k=1

Then we can write F = q B v sinϴ or, F = q( x )

Here the direction of force is given by Fleming’s left hand rule or right hand screw rule .

Fleming’s left hand rule –

According to this rule when we stretch our left hand’s fore finger, middle finger and thumb such that they are perpendicular to each other , if fore finger shows the direction of field, middle finger shows the direction of current( +ve charge) then thumbs gives the direction of force .

Definition of B (magnetic field intensity) –

As we have seen in the equation F = q B v sinϴ ;

If q= 1C , v=1m/s ϴ=90⁰ i.e. sinϴ = 1 then   B= F ;

So we can define magnetic field intensity at a point is equal to the force experienced by a unit charge moving with a unit velocity perpendicular to the direction of magnetic field at that point .

Unit of magnetic field ‘B’ –

From the equation F = q B v sinϴ ,

B = F/qv sinϴ ; then unit of B is NA^-1m^-1 = Tesla (T) ,

And dimension of B is [M A^-1 T^-2]

July 19, 2020
Magnetic effect of current,Oersted experiment and Amperes swimming rule

Magnetic effect of current, Oersted experiment and Amperes swimming rule –

In this topic we will discuss about magnetic effect of current,Oersted experiment and Ampere’s swimming rule.

Magnetic effect of current–

When electric current passes through a conductor (conducting wire) then magnetic field developed around the conductor.

The intimate relationship between electricity and magnetism was discovered 200 years ago , Oersted discovered in the year 1820 that a straight wire carrying current cause a deflection in a nearby magnetic compass needle.

Oersted’s experiment and Ampere’s swimming rule-

According to Oersted’s experiment , we take a needle NS which is free to rotate . We place the needle above a current carrying wire AB . If current flows through the wire in the direction A to B and from B to A then deflection in the needle shown in the figure . Since magnetic needle can deflect only in the interaction of the another magnetic field . so it is confirmed that due to flow of current in a conductor magnetic field setup across around the wire.

The direction of deflection in the magnetic needle due to current in the wire is given by Ampere swimming rule . According to this rule ‘ If a man swimming along the wire in the direction of current with his face turned always towards the needle , so that the current enters through his feet and leaves in his had , then north pole of the magnetic needle will be deflected towards his left hand’ .

July 19, 2020
Circular motion
Circular motion

In this topic we will discuss about what is circular motion, angular displacement,angular velocity, angular acceleration, relation between angular velocity- linear velocity ,and angular acceleration-linear acceleration . Also we will discuss about centripetal acceleration.

Circular motion-

When during the motion body moves on a circular path then the such kind of motion of the body is called circular motion .

Some important topics of circular motion:-
1. Angular displacement – When a body moves on a circular path then the angle traced out by the radius vector at the axis of the circular path in a given time is called angular displacement .
Suppose a body is moving on a circular path of radius ‘r’ , in anticlockwise direction in the plane of paper , with center ‘O’ . let the position of the object changes from P to Q in time t . let <POQ = ϴ , as shown in figure .

Since, angle = arc/radius

So we can write ϴ=PQ/r .

Angular displacement is a vector quantity . Its unit is radian and it has no dimension.
1. Angular velocity – When a body moves on a circular path then angular velocity is defined as the rate of change of its angular displacement . it is denoted by ‘ω’ (omega) .
Its unit is radian/second , and its dimension is [M⁰L⁰T^-1]. It is a vector quantity.

Suppose a point object moving along a circular path of radius r and centre O . let object moves from P to Q in time dt . and <POQ = dϴ .

Then angular velocity ω= dϴ/dt .

Relation between linear velocity and angular velocity –

Suppose an object is moving with uniform angular velocity ω and linear velocity v on a circular path of radius r with centre O .

Object is at P at time t and after time Δt it reaches at Q . <AOP=Δϴ The length of PQ=Δl

Therefore v= Δl/Δt or, Δl=vΔt

But , angular velocity ω=Δϴ/Δt or, Δϴ=ωΔt ,

As we know angle = arc/redius

So, Δϴ=Δl/r or ωΔt=vΔt/r or v=ωr ;

Angular acceleration –

In a circular motion angular acceleration is defined as the time rate of change of its angular velocity .

it is denoted by α . Its unit is rad. S^-2 . And its dimension is [M⁰ L⁰ T^-2].

Relation between linear velocity and angular acceleration – as we know v=ωr

So angular acceleration α= dω/dt =d(v/r)/dt = dv/rdt= a/r ( a= angular acceleration a= dv/dt)

So , a= αr ;

Centripetal acceleration –

In a uniform circular motion , the velocity vector of the object is changing with time . it indicates that the uniform circular motion is the example of circular motion.

Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration . It always acts along the radius and towards the center.

Suppose a particle of mass m moving with a constant speed v and uniform angular velocity ω around a circular path of radius r .

Let at time t the point is at p and after time Δt it reaches at Q .

Here OP=r₁ and OQ = r₂ and , <POQ = Δϴ

Here, angular velocity ω=Δϴ/Δt ;

Let v₁ and v₂ are the velocity at position P and Q respectively . as shown in figure (a) .

Here magnitude of PA and QB are equal which is equal to v.

To find the change in velocity in time interval Δt . we draw p’A’ and P’B’ respectively the velocity vector v₁ and v₂ as shown in above figure (b) .

Here A’B’ = Δv ,

From figure (b)

Δϴ=A’B’/P’A’ = Δv/v

ω Δt= Δv/v ;

ω v = Δv/Δt ;

(v/r)v = a_c

a_c = v²/r =ω²r
July 18, 2020
Complete assignment of electric current
Complete assignment of electric current –

physics classes by Nayan jha already provided assignment of electric current . But this is the Complete assignment of electric current . In students can solve all kind of questions like objective,fill in the blanks,very short, short , long answer type , numericals and hots questions . This assignment will help the students to prepare for 12th board and all kind of competitive examination like JEE and NEET .Specially the objective questions are very helpful for board and competitive examinations.

Before to solve this assignment students must learn complete assignment of electric current .

To download the notes of electric current click here-

Some questions of this assignment is given here-
1. For which of the following dependence of drift velocity ʋ on electric field E is ohm’s law obeyed? (a)ʋ ꭀ E (b) ʋ = constant (c) ʋ ꭀ E^1/2 (d) ʋ ꭀ E²
Ans-a
1. in metals the time of relaxation of electrons:
- (a)increases with increasing temperature
- (b)decreases with increasing temperature
- (c)does not depend on temperature
- (d)changes suddenly at 400 K
Ans-b
1. Constantan wire is used for making standard resistance because it has:
- (a)low specific resistance
- (b)high specific resistance
- (c)negligible temperature coefficient of resistance
- (d)high melting point
Ans-c
1. A potential difference is applied across the ends of a metallic wire. If the potential difference is doubled, the drift velocity:
- (a)will be doubled (c) will be quadrupled
- (b)will be halved (d) will remain unchanged
Ans-a
1. The maximum current that flows in the fuse wire, before it blow out, varies with the radius r as:
- (a)r^3/2 (b) r (c) r^2/3 (d) r^1/2
Ans-a
1. A wire of resistance R is cut into n equal parts. These parts are then connected in parallel. The equivalent resistance of the combination will be:
- (a)nR (b) R/n (c) n/R (d) R/n²
Ans-d
1. The masses of three wires of copper are in the ratio of 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratio of their electrical resistances is:
- (a)1 : 3 : 5 (b) 5 : 3 : 1 (c) 1 : 15 : 125 (d) 125 : 15 : 1
Ans-d
1. Kirchhoff’s first law is based on the law of conservation of:
- (a) charge (b) energy (c) momentum (d) sum of mass and energy
Ans-a
1. Kirchhoff’s second law is based on the law of conservation of:
- (a)charge (b) energy (c) momentum (d) sum of mass and energy
Ans-b
1. A call of emf E is connected across a resistance r. The potential difference between the terminal of the cell is found to be V. The internal resistance of the cell must be:
- (a) 2(E – V)V/r (b) 2(E – V)r/E (c) (E – V)r/V (d) (E – V)r
Ans-c

PDF of the Complete assignment of electric current is given below-

To download the Complete assignment of electric current click here-
July 16, 2020
Complete notes of current electricity

Complete notes of current electricity –

Physics classes by Nayan jha providing Complete notes of current electricity . As you know i am already provided the notes of all topics related to current electricity. But for the students to get the Complete notes of current electricity , here students can find each topics of current electricity at one page .

1. Electric current,

2. Drift velocity , mobility , relation between drift velocity and current

3. proof of Ohm’s law on the basis of drift velocity .

4. Current density , resistance , conductance , resistivity, conductivity , relation between current density and drift velocity.

4. Effect of temperature on resistance and resistivity , graph of resistance(resistivity) Vs temperature for conductors/alloy /semiconductors .

5. Ohmic and non-ohmic conductors ( graph) ,

6.Carbon resister and color coding .

7. Combination of resistors ( series and parallel ) .

8. Internal resistance (r) , terminal potential (V) and emf (E) of the cell and relation between them . also graph between ( V,E, r and R) .

9.Combination of cells (series, parallel ,and mixed grouping) .

10. Kirchhoff’s law (junction and loop rules) ,

11.Wheat -stone bridge .

12. Meter bridge.

13. Potentiometer and its applications ( effects on null points on different -different conditions ).

14. Electrical energy and power .

July 12, 2020
Potentiometer
Potentiometer

In this topic we will discuss about Potentiometer,its working principle and its applications. Also we will know sensitivity of potentiometer.

Before to know about this topic student must read the following topics –

i.- Kirchhoff’s law,to read this topic click here-

ii.- Wheatstone bridge,to read this topic click here-

iii.-Mete bridge,to read this topic click here-

Potentiometer – An instrument used to measure accurately the EMF or potential difference is called potentiometer.

Principle– Its working based on the principle that the potential drop across a conductor is directly proportional to the length of the conductor , if constant current passes through the conductor having uniform area of cross-section .

Construction –

As we know ,

potential V=IR

V = (ρl/A) I

Since current is constant and area of cross-section is uniform so V α l

Or V = K l

K = V/l ( K is a constant called potential gradient)

[ calculation of potential gradient ( students don’t have to derive it in case of potentiometer it is useful to solve the numerical questions) – if Rh = r and resistance of the wire is r and E is the emf of driver cell then , current through the circuit is , i = E/ (R+r)

There fore potential difference at the end the wire is v= i r

V= {E/ (R+r) } r . so potential gradient K = V/l = {E/ (R+r) } r/l. ]

Application of potentiometer –
1. Determination of potential difference using potentiometer –
All the arrangements are shown in the figure-

Let key K is closed then current i starts flowing through the circuit , if R_AB is the resistance of the wire then , We can write current i = ε/R+R_AB ;

Now we close the key K₁ also , in that case potential across the R₁ is ‘V’ and null point J is at length ‘l’ as shown in figure .

Then V = Kl ;

But , K = V_AB/ L so we can write V =( V_AB/ L)l [L Is the total length of potentiometer wire]

Here V_AB = i R_AB = (ε/R+R_AB)R_AB ;

Therefore V = [ {ε/(R+R_AB)}l R_AB]/L ;
1. Comparison of emf of two cells –
A battery of emf ε is connected between the terminals A and B with variable resistance R_h , Now we have to compare the emf ε₁ and ε₂ . We arrange the two cells having emf ε₁ and ε₂ as shown in figure . At first we connect the terminals 1 and 3 , let at that time J₁ is the position where we grt null point , let length of A J₁ is l₁ so we can write ε₁= Kl₁ ……….(i)

Again 1 and 3 is de-attached and terminal 2 and 3 is connected now this time we get null point at J₂ , let AJ₂= l₂ . then we can write ε₂= Kl₂ ………………….(ii)

Dividing equation (i) by equation (ii) we get , ε_1/ ε₂= l₁/l₂ …………..req. eq.

Precautions of experiment- (i) Current through the circuit must be constant .

(ii) the emf of the driving cell (ε) must be greater than emf of the cells whose value has to be compared .
1. Determination of the internal resistance of the cell –
To find the internal resistance r of the cell emf ‘ε’ we setup the circuit as shown in figure.

Initially we close the switch ‘K’ and switch K₂ is remain open .

In that case null point is at J₁ , let AJ₁=l₁

Then we can write ε=Kl₁………………(i)

Now the switch K₁ is closed then the potential across the resistance R is   ‘V’ . And in that case we get the null point at J₂ . And AJ₂=l₂

Then we can write V = Kl₂   ………….(ii)

Dividing (i) by (ii) we get ε/V = l₁/l₂ ……………(iii)

But we know that internal resistance of the cell   r = (ε-V /V)R = [ε/V-1]R ……….(iv)

And from equation (iii) and (iv) we get , r = [l₁/l₂ -1]R ………Req. Eq.

Sensitivity of potentiometer –

The sensitiveness of the potentiometer means the smallest potential difference which can be measured with the help of potentiometer .

The sensitiveness of the potentiometer can be increased ;

(i)by increasing the length of the wire of the potentiometer ,

(ii) by decreasing the value of potential gradient using reducing the current by increasing the resistance .

Difference between EMF of a cell and potential difference –

(i) EMF -It is related with source of current

P.D. – It is related with any two points of the closed circuit.

(ii) EMF – It cause of potential difference between two point of circuit .

P.D. – It is an effect .

(III) EMF – It is the amount of work done in taking a unit charge once round the complete circuit .

P.D. – It is the amount of work done to bring unit charge from one point to another point .

(iv) EMF – It is equal to sum of potential difference across all components of the circuit ( Including all          the cells connected in the circuit .

P.D. – In a closed circuit the different components of the circuit have different potential difference .

(v)   EMF – It is independent of the external resistance and current drawn from the circuit .

P.D. – It depends on the resistance between two points of circuit and current flowing through it .
June 26, 2020
Meter bridge or slide wire bridge
Meter bridge or slide wire bridge

In this topic we will discuss about Meter bridge or slide wire bridge . Principle of Meter bridge or slide wire bridge , its working and, end errors with correction of end error.

Before to know this topic students must read the following topics.

i. Kirchhoff’s rule , To get the notes of this topic click here-

ii. Wheatstone bridge , To get the notes of this topic click here-

Meter bridge or slide wire bridge –

It is a device used to find the value of unknown resistance . It works on the principle of wheat-stone bridge .

It consist a wire of Constantan or Manganin of one meter length and having uniform area of cross-section . As shown in figure wire is stretched between AC , a cell with a key ‘K’ a resistance box R( known resistance ) is attached and an unknown resistance S , in two gap as shown in figure .

When switch ‘k’ is on , we start sliding the jockey from A to c , we notice there is a point where galvanometer shows no deflection ( known as null point) . Let this point is at B , let AB = l

so , BC = 100-l

here R_AB=ƍl/A ( Where A is the area of cross-section)

R_BC = ƍ(100-l)/A

According to wheat-stone bridge principle (P/Q=R/S) We can write

If meter bridge is balanced then R/R_AB = S/R_BC

Or , R/ ( ƍl/A) = S / ƍ(100-l)/A

Or, R/l =S/(100-l) so S = R(100-l)/l ……………eq.

To see the video of meter bridge click on the link given below-

Error end in the meter bridge- cause of the zero end error is following-

(a) The resistance of the copper strip and connecting wire is not taken in account .

(b)The zero mark of the scale may not start from the position where copper strip leaves the wire and 100 cm mark may not on the position where the bridge wire just touches the copper strip .

Processes to remove the end error –

By changing the ‘R’ with ‘S’ and ‘S’ with ‘R’ and repeating the experiment . The average value of S determined earlier and now is free from end error .

*** NOTES- 1. Null point ( where galvanometer shows no deflection) should be at the middle of the bridge wire . Because it becomes more sensitive when all four resistances are nearly same.

2.The meter bridge can’t used to measure very high or very low resistance .
1. the balanced point not effected after interchanging the position of galvanometer and battery .
June 25, 2020