LECTURE 3 : CLASS XII: PHYSICS : ELECTRIC FIELD

CLASS XII   |    PHYSICS    |    CHAPTER 1

      notes prepared by subhankar Karmakar

(the physical quantity written in bold letters are vectors )

ELECTRIC FIELD: 

The electric field or electric intensity or the electric field strength E at a point is defined as the force experienced by a unit positive test charge placed at that point, without disturbing the position of source charge. 

• Electric field E is a vector quantity. 

• SI unit of electric field is N/C. 

• Dimension of electric field is

 [E] = force/charge = [MLT⁻²]/[AT] = [MLT⁻³A⁻¹]

Electric field is an example of vector field:

As the value of electric field E is different at different point, so we can say each point having a position vector r , therefore, the vector E is a function of position vector of a point. Hence, we can say electric field is an example of vector field. 

Principle of superposition of electric fields:

It says the electric field at any point due to a group of charges is equal to the the vector sum of the electric field produced by the each charge individually at that point, when all other charges are assume to be absent. 

LECTURE 1 : CLASS XII: PHYSICS : ALTERNATING CURRENT (AC)

CLASS XII   |    PHYSICS    |    CHAPTER 1

      notes prepared by subhankar Karmakar

Alternating current:

An alternating current is that current whose magnitude changes continuously with time and direction reverses periodically. 

When a coil is rotated in a magnetic field, an alternating EMF is induced in it which is given by the relation

ℰ = ℰₒ sin (ⲱt) ---------------------------- (i)

When applied to a circuit of resistance R, it will produce a current Ｉ such that

Ｉ = ℰ/R = (ℰₒ/R) sin (ⲱt) = Ｉₒ sin (ⲱt)--------(ii)

Therefore, the current in the circuit varies sinusoidally with time and it is called alternating current. 

Ｉ = Instantaneous Current

Ｉₒ = ℰₒ/R = Maximum or peak value of current. It is also called current amplitude. 

Amplitude: The maximum value attained by an alternating current in either direction is called its amplitude or peak value and is denoted by Ｉₒ. 

Time Period: The time taken by an alternating current to complete one cycle of its variations is called its time period and is denoted by T.

This time is equal to the time taken by the coil to complete one rotation in the magnetic field. 

T = 2π/ⲱ ; where ⲱ = angular velocity of the coil. 

Frequency: The number of cycles completed per second by an alternating current is called its frequency and is denoted by f. It is equal to the frequency of rotation of the coil in the magnetic field. 

f = 1/T = ⲱ/(2π)

∴ Ｉ = Ｉₒ sin (ⲱt) = Ｉₒ sin (2πft) = Ｉₒ sin (2πt/T) -------- (iii)

• In India AC supply has a frequency of 50 Hz

Mean / Average Value of AC over a half cycle:

Average value of AC : It is defined as that value of direct current which sends the same charge in a circuit in the same time as is sent by the given alternating current in its half time period. It is denoted by Ｉₘ or Ｉₐᵥ.

LECTURE 2 : CLASS XII: PHYSICS : ELECTRIC CHARGE & FIELD NUMERICALS

CLASS XII   |    PHYSICS    |    CHAPTER 1

      notes prepared by subhankar Karmakar

Numericals on quantization of charge:

1. Which is bigger 1 coulomb or a charge on an electron? How many electronic charges form one coulomb of charge?

2. If a body gives out 10⁹ electrons every second, how much time is required to get a total charge of 1 coulomb from it?

3. How much positive and negative charge is there in a cup of water? (Take the mass of water contained in a cup is 250 g).

4. Calculate the charge carried by 12.5 x 10⁸ electrons. 

5. Estimate the total number of electrons present in hundred gram of water. How much is the total negative charge carried by these electrons? (Take Avogadro number = 6.02 x 10²³ and molecular mass of water = 18).

Numericals on Coulomb's law of electrostatics:

6. Two particles, each having a mass of 5 gram and charge 0.1 microcoulomb , stay in limiting equilibrium on a horizontal table with a separation of 10 centimetre between them. The coefficient of friction between each particle and the table is same. Find the coefficient of friction.

7. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10⁻⁷ C ? The radii of A and B are negligible compared to the distance of separation. What will be the force of repulsion if the two spheres are placed in water?

8. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally remove from both. What is the new force of repulsion between A and B?

9. Two similarly equally charged identical metal spheres A and B repel each other with a force of 2.0 x 10⁻⁵ N. A third identical unchaged sphere C is touched to A, then placed at the mid point between A and B. Calculate the net electrostatic force on C. 

10. Two identical charges, Q each, are kept at a distance r from each other. A third charge q is placed on the line joining the above two charges such that all the three charges are in equilibrium. What is the magnitude, sign and position of the charge q?

Numericals on the superposition principle:

11. Consider three charges q₁, q₂, q₃ each equal to q at the verices of an equilateral triangle of side l. What is the force on a charge Q placed at the centroid of the triangle?

12. Consider the charges +q, +q and -q placed at the vertices of an equilateral triangle. What is the force on a charge?

13. Four equal point charges each 16 μC are placed on the four corners of a square of side 0.2 m. Calculate the force on any one of the charges. 

1 mark questions:

1. Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge q and other an equal negative charge. Are their masses after charging equal?

2. Can two like charges attract each other? If yes, how?

3. Electrostatic experiments do not work well on humid days.  give reasons. 

4. A comb runs through one's dry hair attract small bits of paper. Why? What happens if the hair is wet or if it is a rainy day?

5. Ordinary rubber is an insulator. But the special rubber tyres of aircrafts are made slightly conducting. Why is this necessary?

6. Vehicles carrying inflammable materials is usually have metallic ropes touching the ground during motion. Why?

7. An inflated balloon is charged by rubbing with fur. Will it stick readily to a conducting wall or to an insulating wall? Give reason.

8. What does q₁ + q₂ = 0 signify in electrostatics?

9. Can a body have a charge of 2.4 x 10⁻¹⁹ C? Justify your answer by comment?

10. If the distance between two equal point charges is doubled and their individual charges are also doubled, what would happen to the force between them?

MOVING COIL GALVANOMETER

Galvanometer: 

A galvanometer is a device to detect current in a circuit. 

Principle: 

A current carrying coil placed in a magnetic field experiences a current dependent torque, which tends to rotate the coil and produces angular deflection. 

Construction:

A galvanometer consists f a rectangular coil of fine insulated copper wire wound on a light non magnetic metallic frame. The two ends of the axle of this frame are pivoted between two bearings. The motion of the coil is controlled by a pair of hair springs of phosphor bronze. The inner ends of the springs are soldered to the two ends of the coil and the outer ends are connected to the the binding screws. The springs provide the restoring torque and serve as current leads. A light aluminium pointer attached to the coil measures its deflection on a suitable scale. 

The coil is symmetrically placed between the cylindrical pole pieces of strong permanent horseshoe magnet. 

Theory and Working:

Let Ｉ = current flowing through the coil PQRS

     a, b = sides of the coil PQRS

       A  = ab = area of the coil

       θ = angle between the direction of B and normal to the plane of the coil.

       N = number of turns in the coil

Since the field is radial, the plane of the coil always remains parallel to the field B. Magnetic forces on the sides QR and SP are equal, opposite and collinear, so their resultant is zero. According to Fleming's left hand rule, the side PQ experiences a normal inward force equal to NＩbB why is the side QR experiences an equal normal out what force. The two forces on sides PQ and RS are equal and opposite. They form a couple and exert a torque given by 

τ = one of the force x perpendicular distance between them 

τ = F a sin θ = ＩbBa sin θ = ＩBA sin θ

[ ∵ ab = A]

If the rectangular loop has N turns, the torque increases N times ie.,

τ = NＩBA sin 90° = NＩBA

Here, θ = 90°, because the normal to the plane of coil remains perpendicular to the field B in all positions. 

The torque τ deflects the coil through an angle α. A restoring torque is set up in the coil due to elasticity of the springs such that

      τᵣ ∝ α   or   τᵣ = kα 

Where K is is the the torsion constant of the springs. 

Restoring Torque = Deflecting Torque

kα = NＩBA

Or  α = (NBA/k)Ｉ

Or      α ∝ Ｉ

Thus the deflection produced in the galvanometer coil is proportional to the current flowing through it. Consequently, the instrument can be provided with a scale with equal divisions along a circular scale to indicate equal steps in current. Such a scale is called linear scale.

Ｉ = (k/NBA) α = Ｉ = Gα

G = (k/NBA) is constant for a galvanometer and is called galvanometer constant for current reduction factor of the galvanometer.

TORQUE EXPERIENCED BY A CURRENT LOOP IN A UNIFORM MAGNETIC FIELD

Torque on a current loop in a uniform magnetic field:

A rectangular coil PQRS suspended in a uniform magnetic field B. The axis of the rectangular coil is perpendicular to the field. 

Let Ｉ = current flowing through the coil PQRS

     a, b = sides of the coil PQRS

       A  = ab = area of the coil

       θ = angle between the direction of B and normal to the plane of the coil.

 

Direction of the area and B makes an angle θ

all the forces acting on the sides of the rectangular coil PQRS

According to Fleming's left hand rule, 

i) the magnetic force on the side QR is F₁

 and it is acting upward.

F₁ = Ｉ(a xB) = ＩaB

ii) the magnetic force on the side SP is F'₁ and it is acting downward. 

F'₁ = Ｉ(a xB) = ＩaB

So, net force along vertical direction is zero as 

F₁ and F'₁ are equal and opposite as both are acting along the axis of the coil.

iii) the magnetic force on the side SR is F and it is coming out of the board.

F = Ｉ(b xB) = ＩbB

iv) the magnetic force on the side QP is F' and it is going into the board.

F' = Ｉ(b xB) = ＩbB

coil as seen from the top : m is the direction of the magnetic moment as well as coil area (perpendicular to the plane of the coil).

Therefore F and F' will produce a torque τ

We know, 

τ = one of the force x perpendicular distance between them 

τ = F a sin θ = ＩbBa sin θ = ＩBA sin θ

[ ∵ ab = A]

If the rectangular loop has N turns, the torque increases N times ie.,

τ = NＩBA sin θ

But there is one physical quantity called "magnetic moment" or m = NＩA

τ = m B sin θ = m x B

The direction of the torque τ is such that it rotates the loop clockwise about the axis of the loop. 

The torque will be zero when θ = 0 ie., When the plane of the loop is perpendicular to the magnetic field. 

The torque will be maximum when θ = π/2 and τ = NＩBA ie., when the plane of the loop is parallel to the magnetic field. 

Lecture- 5 : CLASS-X: SCIENCE : Chapter: REFLECTION OF LIGHT & NUMERICALS

CLASS X   |    SCIENCE    |    LIGHT

      notes prepared by subhankar Karmakar

                                                                         

Numericals on concave mirror:

Q1. What is the nature of a mirror having a focal length of, + 4 cm?

1. Ans. As the focal length is positive, the mirror is convex mirror. 

Q2. What kind of mirror can have a focal length of, - 6 cm?

2. Ans. As the focal length is negative, the mirror is a concave mirror. 

Q3. If the radius of curvature of a mirror is - 20 cm, what will be its focal length? What type of mirror it is? 

3. Ans. As the focal length is half of the the radius of curvature, so, f = -10 cm. As the  focal length is negative it is a concave mirror. 

Q4. Focal length of a small concave mirror is 2.5 cm. In order to use this concave mirror as a dentist's mirror, what must be the distance of tooth from the mirror?

4. Ans. As a dentist's mirror needs a real and magnified image of the tooth, the tooth must be placed in between pole and focus. Therefore, the distance of the tooth must be less than focal length of 2.5 cm. 

Q5. We wish to obtain an erect image of an object using a concave mirror of focal length 15 cm. 

a. What should be the range of distance of the object from the mirror?

b. What is the nature of the image?

c. Draw a ray diagram to show the image formation in this case. 

5. Ans. a. In order to obtain an erect image of an object with a concave mirror, the object should be at a distance less than its focal length. Therefore, here, the object must be placed at a distance less than 15 cm. 

        b. The nature of the image will be virtual. The image will be larger than the object

Q6. The image formed by a concave mirror is seen to be virtual, erect and larger than the object. Where should we place the object?

6. Ans. We should place the object in between pole and focus of the mirror. 

Q7. A concave mirror has a focal length of 20 cm. Where should an object be placed in front of this Concave mirror so as to obtain an image which is real, inverted and same size of the object?

7. Ans. When the object is placed at the centre of curvature, it produces an image, which is real, inverted and same size of the object. The distance of the centre of curvature from the pole is called radius of curvature and it is equal to twice of the focal length. 

Hence, the object must be placed at a distance (2x20 = 40 cm) 40 cm from the pole in front of the concave mirror. 

Q8. An object is placed in front of a concave mirror focal length 10 cm. Find the object distance if it produces a magnified real image?

8. Ans. If the object is placed in between focus and the centre of curvature,  then the image produce is real and inverted and magnified. Therefore, the object must be in between focus (f) and centre of curvature (2f). 

So, the object must be placed in between 10 cm to 20 cm from the pole in front of the mirror. 

Q9. An object is placed in front of a concave mirror focal length 5 cm. Find the object distance if it produces a diminished real image?

9. Ans. When the object is beyond the centre of curvature of the concave mirror it produces a diminished real image. Therefore the object must be at least more than 2f = 2x5 cm = 10 cm from the pole in front of the mirror.

Q10. An object is 20 mm in front of a concave mirror which produces an upright image or erect image. The radius of the curvature of the mirror is

a. Less than 20 mm       b. Exactly 40 mm

c. In between 20 mm and 40 mm

d. More than 40 mm. 

10. Ans. As for an erect image, object must be placed in between pole and focus, then focal length in this case is more than 20 mm. Therefore, the centre of curvature must be more than 2 x 20 mm = 40 mm. 

d. is the correct answer. 

Q11. Find the size, nature and position of image formed when an object of size 2 cm is placed at a distance of 9 cm from a concave mirror of focal length 6 cm. 

11. Ans. Focal length, f = - 6 cm and object distance, u = - 9 cm. Height of the object, h₁ = 2 cm. 

We know, mirror formula, 1/u + 1/v = 1/f

⟹ 1/v = 1/f - 1/u

⟹ 1/v = (u - f)/fu

⟹ v= fu/(u - f)

⟹ v = (-6)(-9)/( - 9 + 6) = (6x9)/(-3) 

∴ v = - 18 cm (position)

Magnification, 

m = (h₂/h₁) = (- v/u) = 18/(-9) = - 2

h₂ = m x h₁ = - 2 x 2 = - 4 cm (inverted)

Position: The image is 18 cm in front of the mirror. 

Nature: The image is real, inverted and magnified.

Size: The image is 4 cm high, inverted and twice magnified and below the principal axis.

Q12. An object 1 cm high is placed at a distance of 10 cm from a concave mirror which produces a real image 2 cm high. (i) what is the focal length of the mirror? (ii) find the position of the image.

Q13. A concave mirror produces three times magnified real image of an object placed at 8 cm in front of it. Where is the image located? What is the focal length of the mirror?

Q14. What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is (i) +2 (ii) - 0.50 ?

Q15. An object is placed at a distance 8 cm from a concave mirror of focal length 12 cm. 

a. Draw a ray diagram for the formation of image.

b. Calculate the image distance.

c. State two characteristics of the image formed. 

Q16. At what distance from a concave mirror of focal length 12 cm should an object 1 cm long be placed in order to get an erect image 4 cm tall?

Q17. When an object is placed at a distance of 4 cm from a concave mirror, its image is formed at 6 cm behind the mirror. Calculate the focal length of the mirror. 

Q18. An object is placed in between principal focus and centre of curvature in front of a concave mirror. Draw a ray diagram to show how the image is formed, and describe its size, position and nature.

Lecture 1: CLASS XI : PHYSICS : CHAPTER - 5 : LAWS OF MOTION

FORCE: 

Force may be defined as an agency (a push or a pull) which changes or tends to change the state of rest or of uniform motion or the direction of motion of a body.

Effects produced by a force:

1. Force can change speed of an object.

When force is applied on a body the body starts to move. Again, when a force exerted by the brakes slows or stops moving train.

2. Force can change the direction of motion of an object.

Force exerted by a bat to a ball, changes the direction of the ball. 

3. Force can change the shape of an object. 

If we apply a force on a rubber ball, round shape of a rubber ball gets distorted.

Galileo's Laws of inertia:

A body moving with certain speed along a straight path will continue to move with same speed along the same straight path in the absence of external forces. 

INERTIA: 

The inherent property of a material body by virtue of which it cannot change, by itself, its state of rest or of uniform motion in a straight line is called inertia. 

Different types of inertia:

a. Inertia of rest: The tendency of a body to remain in its position of rest is called inertia of rest. 

Example: A person standing in a bus falls backward when the bus suddenly starts moving forward. 

b. Inertia of motion: The tendency of a body to remain in its state of uniform motion in a straight line is called inertia of motion. 

Example: When a moving bus suddenly stops, a person sitting in it falls forward. 

c. Inertia of direction: The inability of a body to change by itself its direction of motion is called inertia of direction.

Example: When a bus takes a sharp turn, a person sitting in the bus experiences a force acting away from the centre of the curved path. It is due to inertia of direction. 

Measurement of inertia of a body:

Mass of a body is the measure of its inertia. If a body has more mass, it has more inertia, it means it is more difficult to change its state of rest or of uniform motion. 

Linear momentum (p):

Momentum of a body is the quantity of motion possessed by the body. It is equal to the product of Mass and velocity of the body.

Momentum = mass x velocity 

Momentum is a vector quantity because the velocity v is a vector and mass m is a scalar. Its direction is same as the direction of the velocity of the body. Its magnitude is given by

p = mv

SI unit of momentum = kg m/s

CGS unit of momentum = g cm/s

The dimensional formula of momentum = [MLT⁻¹]

Q1. Two objects, each of mass m and velocities v₁ and v₂. If v₁> v₂, which one has more momentum?

Ans: p₁ = mv₁ and p₂ = mv₂

∴ (p₁/p₂) = (mv₁/mv₂) = (v₁/v₂)

As v₁> v₂ , so p₁> p₂

Q2. Two objects having mass m₁ and m₂ such that m₁> m₂ , and same velocity v. Which one has more momentum?

Ans: p₁ = m₁v and p₂ = m₂v

∴ (p₁/p₂) = (m₁v/m₂v) = (m₁/m₂)

As m₁> m₂ , so p₁> p₂

Q3. Two objects having same momenta (p₁ = p₂), if m₁> m₂, which one has more velocity?

Ans. p₁ = m₁v₁ and p₂ = m₂v₂

As p₁ = p₂

∴ m₁v₁ = m₂v₂ 

or  (v₂/v₁) = (m₁/m₂)

As m₁> m₂ , so v₁< v₂

Velocities of bodies having equal linear momenta are inversely proportional to their masses. 

So, when two objects have equal linear momentum, the lighter object will move faster than the heavier one.