4. Moving Charges and Magnetism

Magnetic Field

Region around a current carrying conductor in which electro-magnetism effect is produced, is said to be Magnetic Field.

It is generally denoted by B. Magnetic field is a vector quantity. The dimension of magnetic field is [MA⁻¹T⁻²]. The SI unit of magnetic field is NA⁻¹m⁻¹ or Tesla.

B = F_m/qvsinθ

Here,
Fm = Magnetic Force
q = Magnitude of the charge
v = Velocity of the charge
θ = Angle between velocity and magnetic force.

Magnetic field vertically upward and downward in a plane are generally denoted by conventional sign (.) and (x) respectively. 

Total Magnetic field from different sources is the vector sum of the different magnetic field.
Therefore,

B = B1 + B2 + B3 + ...................... 
Super position principle is used during the sum of magnetic field.

Lorentz Force

The field in which both electric and magnetic field are in existence, the total force experienced by charge q due to motion is said to be Lorentz Force.

Lorentz Force = Force on charge due to electric field + Force on charge due to Magnetic field
F = Fe + Fm
Here,
F = Lorentz Force
Fe = Electric Force
Fm = Magnetic Force

Therefore,
Lorentz Force = F = qE + qvBsinθ

Special Cases of Magnetic Force

Magnetic Force = Fm = qvBsinθ
Case I
When θ = 0⁰ or 180⁰
Fm = 0
Charge will be moved either parallel or antiparallel of magnetic field.

Case II

When  θ = 90⁰ 
Fm = qvB
Charge will be moved along the perpendicular direction of magnetic field.

Magnetic Force on a current carryin conductor

                 Consider about a conductor which is placed in a magnetic field B along z axis. The direction of the magnetic field is along x axis. So that magnetic force will be along y axis according to Fleming's left hand rule.
                   We know that a large numbers of electrons are present in free state in a conductor which move opposite of current with drift velocity.

Mathematical Calculation

Let the length of the conductor = l
cross - sectional area of the conductor = A
drift velocity = v
current = I
charge = -e
No. of electrons per unit volume of the conductor = n

Now, according to Lorentz Magnetic Force
Fm = -evBsinθ
Now consider a small length dl.

Volume of conductor for this length = Adl
No. of electrons = nAdl
Charge = -enAdl

Magnetic Force for this length 
dFm = -enAdlvBsinθ

drift velocity for dl = v = -dl/dt             (The direction of dl is opposite to drift velocity)

Therefore,
dFm = -enAdlBsinθ.-dl/dt ---------------- (i)

enAdl/dt = I

Now, from (i)

dFm = I(dlxB)

For whole conductor

Fm = IlBsinθ

Direction of Fm, I and B are determined by Fleming's left hand rule.

Special Cases

Case I

When θ = 0⁰ or 180⁰
Charged particles move either parallel or anti parallel of magnetic field.

Case II

When θ = 90⁰ 

Charged particles move along the perpendicular of magnetic field.

Under this circumstances the charged particles move in circular path.

                  Consider about a charged particles of mass m and charge q moves in a circular path of radius r.

Centripetal Force experienced by the particle = mv²/r

Magnetic Force = qvB

qvBsinθ = mv²/r

r = mv/qB

Let the charged particle takes T time to move one rotation.
Therefore,
T = 2πr/v
Now putting the value of r
T = 2πm/qB

Frequency ν = 1/T = qB/2πm

Case III

When θ is other than 0⁰, 90⁰ or 180⁰

Let the angle be θ 

Magnetic field B is along x axis, current is along z axis and magnetic force Fm is along y axis.

Under this circumstances the charged particle moves in hellical path.

Vertical Components of drift velocity of charged particle = vsinθ
Horizontal Components of drift velocity of charged particle = vcosθ

Particle along vertical components moves along circular path.

Centripetal Force = m(vsinθ)²/r
Magnetic Force = qBvsinθ

Centripetal Force = Magnetic Force
r = mvsinθ/qB

T = 2πr/ vsinθ

Now putting the value of r

T = 2πm/qB

Frequency ν = 1/T = qB/2πm

Motion in Combined Electric and Magnetic Field

Consider about a charge q moves in a field in which both electric and magnetic field are perpendicular to each other.

Electric force acts along electric field and magnetic force opposite to electric force. Direction of magnetic field is determined by Fleming's right hand rule.

Mathematical Calculation

Fe = Fb

qE = qvB.sin90⁰
qE = qvB
v = qE/qB
v = E/B

Magnetic Field due to a current element, Biot Savart Law

Biot Savart Law states that 
"The magnitude of the magnetic field is proportional to the current, the element length and inversely proportional to the square the distance of the point from current element at which magnetic field is determined. The direction of the magnetic field is perpendicular to the plane containing element length and distance."

          Consider about a finite conductor XY carrying current I. dl is the current element. There is a point P at a distance r from the current element. At this point magnetic field dB is to be determined. The angle between dl and displacement vector r is θ.

The relevant figure is drawn below

Now according to Biot Savart Law

dB∝ Idlxr/r³
dB = 𝛍ₒIdlxr/4𝛑r³

Here 
𝛍ₒ/4𝛑 is a constant of proportionality ( The medium is vacuum)

Therefore, magnitude of the magnetic field
।dB। = 𝛍ₒIdlsinθ/4𝛑r²

The value of 𝛍ₒ/4𝛑 = 10⁻⁷ Tm/A
𝛍ₒ is the permeability of free space or vacuum.

There are some similarities, as well as differences, between Biot Savart's Law and Coulomb's Law. Which are as follows

1. Both depend inversely on the square of distance from the source of interest.
2.The principle of superposition applies to both magnetic and electric fields.
3. Both magnetic field and electric field is linear to their sources Idl (current element) and q (source of electric charge) respectively.
4. The electric field is produced by scalar source q (electric charge) whereas magnetic field is produced by vector source Idl (current element)
5. The electric field is along the displacement vector whereas magnetic field is perpendicular to the plane containing the displacement vector r and current element Idl.
6. The magnetic field at any point in the direction of dl is zero.

Relation between permitivity of free space (𝛆ₒ) and permeability of free space (𝛍ₒ) and speed of light (c).

𝛆ₒ𝛍ₒ = 4𝛑𝛆ₒ.𝛍ₒ/4𝛑

= 10⁻⁷/9x10⁹
= 1/9x10¹⁶
= 1/(3x10⁸)²
= 1/c²

Chapter 4 Class XII

14. Natural Resources Class IX NCERT Text Book

Q.No. 1. How is our atmosphere different from the atmospheres on Venus and Mars ?

Answer:

Atmosphere of Earth has 79% nitrogen, 20% oxygen and a small amount of carbon dioxide, water vapour and other gases.
This existence of life is possible on Earth.

Whereas atmospheres of Venus and Mars have 95% - 97% carbon dioxide.
Existence of life is not possible.

Q.No. 2. How does the atmosphere act as a blanket ?

Answer:

Air is bad conductor of heat. The atmosphere keeps the average temperature of the Earth fairly constant during the day and even during the course of the whole year such as blanket.

The atmosphere prevents the sudden increase and decrease of temperature during the daylight hours and the night just like blanket.

The atmosphere slows down the escape of heat into outer space. 

Due to following behavior of atmosphere acts as blanket.

Q.No. 3. What causes winds ?

Answer:

The movement of air causes winds. When air is heated by radiation from the heated land or water, it rises. Land gets heated faster than water. A region of low pressure is created. The movement of air from one region to the other region creates winds.

The rotation of Earth and presence of mountain ranges cause winds.

Q.No. 4. How are clouds formed ?

Answer

Water vapour moves from high pressure region to low pressure region.
A large amount of water evaporates and goes into air due to heating of water bodies. The hot air rises up carrying the water vapour with it. As the air rises, it expands and cools. This cooling causes the water vapour in the air to condense in the form of droplets. 
This process forms the clouds.

Q.No. 5. List any three human activities that you think would lead to air pollution.

Answer

There are following human activities which lead to air pollution.

1. Burning of coal and petroleum.
2. Deforestation
3. Industrialization

Q.No. 6. Why do organisms need water ?

Answer

There are following reasons due to which organisms need water.

1. Usually, the reactions that take place in our body or within cells occur between substances that are dissolved in water. So, all cellular process need water.

2. Most of the substances are transported in a dissolved form, water is necessary.

Q.No. 7. What is the major source of water in the city/town/village where you live ?

Answer

River is the major source of fresh water.

Q.No. 8. Do you know any activity which may be polluting this water source ?

Answer

The discharge of waste water from homes, industries, hospitals, etc. into the river pollutes this fresh water source.

Class IX NCERT Text Book Questions Answers

Motion of a car on a banked road

               Consider about a car of mass m kg moves on banked road of inclination θ. During the motion the car experience circular motion.

Mathematical Calculation

There is no acceleration along the vertical direction. The net force along this direction must be zero.

N cosθ = mg + f sinθ ------------ (i)

The horizontal components of N and f provide the centripetal force.

N cosθ + f cosθ = mv²/r ---------- (ii)

The frictional force 

f ≤ μ_s N
For maximum velocity v

f = μ_s N
Now putting the value of f in equation no. (i)

N cosθ = mg +  μ_s N sinθ
Therefore,
N = mg/(cosθ  - μ_s sinθ)

Again putting the value of f in equation no. (ii)

N sinθ + μ_s N cosθ = mv²/r

Now substitute the value of N

mg( sinθ  + μ_s  cosθ)/(cosθ  - μ_s sinθ) = mv²/r

Therefore,

v = {rg( sinθ  + μ_s  cosθ)/(cosθ  - μ_s sinθ)}^1/2

This is maximum velocity of a car on a banked road.

Heights And Distances

Chapter 9

Class X

Exercise 9.1

Q.No. 1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30⁰

Solution:

From question it is clear that
AC = 20 m.
<ACB = 30⁰ = θ 
AB = ?

We know that

Sinθ = AB/AC

Now, putting the value of θ, AB, AC.

Sin30⁰  = AB/20
1/2 = AB/20
∴ AB = 10m Ans.

Q.No. 2. A tree breaks due to storm and broken parts bends so that the top of the tree touches the ground making an angle 30⁰ with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

From question it is clear that

Height of the tree = AB + AC
θ = 30⁰
BC = 8 m.

From figure 

Cos 30⁰ = BC/AC
√3/2 = 8/AC
∴ AC = 16/√3

Again, 

Tan 30⁰ = AB/BC
1/√3 = AB/8

∴ AB = 8/√3

Height of the tree = AB + AC 

= 8/√3 + 16/√3

= 24/√3 
= 8√3
= 8x1.732
= 13.856 m Ans.

Q.No. 3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30⁰ to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60⁰ to ground. What should be the length of the slide in each case?

Solution:

From question it is clear that
For first slide
AB = 1.5 m
α = 30⁰
Length of slide = AC = ?

For second slide
PQ = 3 m
β = 60⁰
Length of slide = PR = ?

For first slide
According to application of Trigonometry
Length of slide = AC = AB/Sin30⁰
∴ AC = 1.5x2 = 3 m

Now, for second slide
Length of slide PR = PQ/Sin60⁰
∴ PR = 3x2/√3
PR = 2√3
PR = 2x1.732
PR = 3.464 m 

Therefore length of slides are
3m and 3.464 m Ans.

Q.No. 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30⁰. Find the height of the tower.

Solution:

From question it is clear that
Let the height of the tower  = AB = ?
BC = 30 m
Angle of elevation = 30⁰

According to application of Trigonometry
AB = BCxTan30⁰
∴ AB = 30/√3
AB = 10√3
Therefore, height of the tower = 17.32 m Ans.

Q.No. 5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60⁰. Find the length of the string, assuming that there is no slack in the string.

Solution:

Let the length of the string be AC = x which represents the hypotenuse.

The inclination represents the angle of elevation = 60⁰

The height of the kite from ground = AB = 60 m.

The concerned right angle triangle is drawn below

From figure it is clear that

Sin60⁰ = AB/AC

Now, putting the value of AB and AC.

√3/2 = 60/x

∴ x = 120/√3

= 40√3 m Ans.

Q.No. 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30⁰ to 60⁰ as he walks towards the building. Find the distance he walked towards the building.

Solution:

Let the distance he walked towards the building be CD = x m.

The angle of elevation for Triangle ABC = 30⁰

The angle of elevation for Triangle ABD = 60⁰

Height of the building from the eye of the boy = AB = 30 - 1.5 = 28.5 m.

The concerned right angle triangle is drawn below

Consider in right angle triangle ABC
Tan30⁰ = AB/BC
Now, putting the value of BC
BC = 28.5.√3

Now consider in right angle triangle ABD
Tan60⁰ = AB/BD
BD = AB/Tan60⁰
∴ BD = 28.5/√3 = 9.5.√3 m.
Therefore,
CD = BC - BD = 28.5.√3  - 9.5.√3 = 19√3 m Ans.