Application of Derivatives

Exercise 6.1 Class 12

Application of Derivatives

Mathematics

Class 12

Exercise 6.1

Solution of

QNo. 3, 4 and 5

4.7 Ampere's Circuital Law

Subject: Physics

Class XII

Chapter: 4. Moving Charges and Magnetism

Ampere's Law

 "The integral of magnetic field around a closed path is equal to the product of absolute permeability and total current passing through it."

Mathematical form

∮B.dl = μ₀.I

Here,

B = Magnetic field

dl = Current element

μ₀ = Absolute Permeability

I = Current passing through it

Proof

Consider about an infinitesimal carrying conductor through which I current passing is passing. Magnetic field lines are produced around the conductor as the form of concentric circles.

Magnetic field at a distance due to this infinitesimal conductor. 

B = μ₀.2I/4πa ................... (i)

Now consider about a circle of radius a having a current element dl.

Both B and dl have same direction because B is along the the tangent of circle.

Hence,

B.dl = Bdlcosθ = Bdlcos0⁰ = Bdl

Now integrate the closed path

∮B.dl = ∮Bdl .............. (ii)

Now putting the value of B

∮B.dl = ∮μ₀.2Idl/4πa

∮B.dl  = μ₀.2I/4πa∮dl

For complete cirecle

∮dl = 2πa

∮B.dl  = μ₀.2I.2πa/4πa

∮B.dl = μ₀.I

Proved

4.6 Magnetic Field on the Axis of Circular Current Loop

Subject: Physics

Class : XII

Chapter 4. Moving Charges and Magnetism

4.6 Magnetic Field on the Axis of Circular Current Loop

Consider about a circular loop of radius R carrying the current I. The loop is placed in the y-z plane with its centre at the origin O. The x axis is the axis of the loop. There is a point P at a distance x from the centre of the loop at which magnetic field is to be determined.

Mathematical Calculation

dl is the conducting element of the loop.

The magnitude of dB of the magnetic field due to dl is according to Biot-Savart Law

dB = 𝝁₀I|dlxr|/4𝜋r³
Now 
r² = x² + R²
Any element to loop will be perpendicular to the displacement vector r from dl to the axial point P is in the x-y plane.
Hence,
|dlxr| = rdl
∴dB = 𝝁₀Idlr/4𝜋r³ ............... (i)
Now putting the value of r
dB = 𝝁₀Idl/4𝜋(x² + R²)
The direction of dB is perpendicular to the plane formed by dl and r.
It has an x- component dBₓ and a component perpendicular to x-axis dB⫡.
When the components perpendicular to the x- axis are summed over, they cancel out.
Only the x component survives.
∴dBₓ = dBcos𝜃
From figure 
cos𝜃 = R/√(x² + R²) ............. (ii)
Now from eqn. no. (i) and (ii)
dBₓ = 𝝁₀IdlR/4𝜋(x² + R²)³/²
For whole circular loop
dl = 2𝜋R
Magnetic field at P due to entire circular loop
= B = Bₓi = 𝝁₀IR²/2𝜋(x² + R²)³/²i
Field at the centre of the loop. Here x = 0
B₀ = 𝝁₀I/2R

The direction of the magnetic field is determined by the help of 
Fleming Right Hand Thumb rule.

4.4.2 Cyclotron Class XII.

The Cyclotron is a machine which is used to accelerate the charged particles.

This machine was invented by E.O. Lowrence and M.S Livingston. 

Construction

It has two hollow chambers of D shaped is called dees. There is some gap between the dees in which source of positive particles are kept. These dees are connected by high frequent oscillator which provides high frequent electric field between the gap of dees. This machine is placed between the two powerful electromagnetic poles.

Working Principle

Positive charged particles accelerated from D1 to D2 when D1 and D2 are positive and negative respectively. In D2 charged particles accelerated perpendicular to the magnetic field.

The schematic figure of Cyclotron is drawn below.

The charged particles move in a semi circular path perpendicular to the magnetic field.

Let the time taken to complete one revolution is T.

T = 2πm/qB
Frequency = 1/T
= qB/2πm

During the circular motion of charged particles.
mv²/r = qvB
v = qBr/m

Kinetic Energy = mv²/2
Now, putting the value of v.
Kinetic Energy = q²B²r²/2m.

Time of revolution is independent of speed of particles and radius of trajectory.

Uses of Cyclotron are as follow

1. Uses in the nuclear reactor plant
2. To implant ions into solids
3. To synthesis materials
4. To produce radio active substances in the hospital.

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