IB Unit 10. Fields: Describing fields and Fields at work

10 | Force Fields

IB Physics Content Guide

Big Ideas

        Opposite charges/poles attract while like charges/poles repel

        The force between charged particles demonstrates the same relationship as the force between bodies with mass

        A force field describes the force at a location per unit mass, charge, or current

        A current flowing through a conductor produces a magnetic field

        The relative directions of current, magnetic field, and electromagnetic force can be found using the right-hand rules

Content Objectives

10.1 – Static Electricity

 

I can explain how objects can become charged

 

 

 

I can qualitatively describe the reactions between charged particles

 

 

 

I can describe the process of grounding a charged object

 

 

 

     

10.2 – Electrostatic and Gravitational Force

 

I can use Coulomb’s Law to relate electrostatic force to particle charge and separation distance

 

 

 

I can use the Law of Gravitation to relate gravitation force to object mass and separation distance

 

 

 

I can determine the units of Coulomb’s Constant and the Gravitation Constant using unit analysis

 

 

 

I can describe how the sign of the calculated electrostatic force indicates attraction or repulsion

 

 

 

I can compare and contrast electrostatic and gravitation forces

 

 

 

I can discuss the impact of permittivity on Coulomb’s Constant

 

 

 

     

10.3 – Force Fields

 

I can calculate force between objects with a net charge or mass

 

 

 

I can draw the vector force field for electric and gravitational fields

 

 

 

I can describe the role of a test charge or test mass in representing force fields

 

 

 

I can describe how the magnitude of a force changes with distance from an object

 

 

 

I can calculate field strength with proper units around a single object

 

 

 

I can calculate the net field strength based on two or more objects

 

 

 

I can determine the location where the net field strength is zero

 

 

 

     

10.4 – Magnetism

 

I can describe the pole conditions required for magnetic attraction and repulsion

 

 

 

I can explain what happens when a dipole magnet is cut into pieces

 

 

 

I can describe the role of magnetic domains in magnetizing and de-magnetizing a material

 

 

 

I can draw in magnetic field lines around a magnet with a north and south pole

 

 

 

I can describe the layout of the Earth’s magnetic poles

 

 

 

     

10.5 – Right Hand Rule and Electromagnets

 

I can use the right-hand rule to draw in a magnetic field around a current carrying wire

 

 

 

I can use the right-hand rule to predict the current direction through a wire with a surrounding field

 

 

 

I can indicate a vector that is pointing into or out of the page

 

 

 

I can use the right-hand rule to predict the location of the north pole in a coil of wires

 

 

 

I can describe some applications of electromagnets in use today

 

 

 

I can describe the design factors that affect the strength of an electromagnet

 

 

 

     

10.6 – Electromagnetic Force

 

I can use the right-hand rule to predict the force direction on a charge moving through a field

 

 

 

I can use the right-hand rule to predict the force direction on a current carrying wire placed a field

 

 

 

I can describe the general functions of electric motors and generators

 

 

 

I can calculate the magnetic field strength and force on a wire or moving charged particle

 

 

 

I can predict the trajectory of a charged particle moving through a magnetic field at different speeds

 

 

 

     

10 | Force Fields

Shelving Guide

Forces between objects

Coulomb’s Law

 

Variable Symbol

Unit

 

Data Booklet Equations:

Electrostatic Force

F

N

 

Object 1 Charge

q1

C

 

Object 2 Charge

q2

C

 

Separation Distance

r

M

 

Coulomb Constant

k

N m2 C-2

 

k = 8.99 × 109 N m2 C-2

Permittivity of Free Space

ε0

C2 N-1 m-2

 

ε0 = 8.85 × 10-12 C2 N-1 m-2

Universal Law of Gravitation

 

Variable Symbol

Unit

 

Data Booklet Equation:

Gravitational Force

F

N

 

Object 1 Mass

M

kg

 

Object 2 Mass

m

kg

 

 

Separation Distance

r

m

 

Gravitational Constant

G

N m2 kg-2

 

G = 6.67 N m2 kg-2

Force Fields

Electric Field

Gravitational Field

Symbol

E

Data Booklet Equation:

Symbol

g

Data Booklet Equation:

Units

F C-1

Units

F kg-1

Magnetic Fields

Right Hand Rule:

Right Hand Rule #1

Right Hand Rule #2

Right Hand Rule #3

 

Magnetic field around a current carrying wire

Pole orientation for a coil of wire (electromagnet, solenoid, etc.)

Electromagnetic force direction on a wire or moving particle

 

Thumb

Current

Thumb

North Pole

Thumb

Current

 

Fingers

Magnetic Field

Fingers

Current

Fingers

Magnetic Field

 

Palm

Force

 

 

 
          

 

Electromagnetic Force

 

Variable Symbol

Unit

 

Data Booklet Equations:

Magnetic Force

F

N

 

 

Magnetic Field Strength

B

T

 

Wire:

Current

I

A

 

Wire Length

L

m

 

 

Angle to Field

θ

°

 

Particle:

Particle Charge

q

C

 

Particle Velocity

v

m s-1

 

 

 Charged Particles Moving through a Magnetic Field

 

Magnetic Field | Out of Screen

Magnetic Field | Into Screen

Positive Particle

Negative Particle

 

10.1 Describing Fields

Gravitational…

Fields

  • Magnitude: g = GM/r², in N kg^-1.

  • Field lines: For a point or spherical mass M, the field is radial, with the field lines towards that mass. In the case of a planet, when very close to its surface, the planet may be considered flat and the field uniform.

approximationperpendicular.png

Potential (Vg)

  • Definition: “The gravitati

  • onal potential at a point P in a gravitational field is the work done per unit mass in bringing a small point mass from infinity to point P”.

  • Vg = W/m = -GM/r, in J kg^-1.

  • Work: the work done depends only on the change of the potential, not on the path taken.

    • Positive work is done on the test object, increasing the gravitational potential.

    • Negative work is done by the test object, decreasing the gravitational potential.

Potential energy (EP)

  • Definition for one body: “The gravitational potential energy of one body is the work done to bring one mass from infinity to a specific point”.

  • Definition for two bodies: “The gravitational potential energy of two bodies is the work done in bringing

  •  the bodies to their present position when they were infinitely apart”.

  • EP = -GMm/r. (negative sign implies that force is attractive and that +GMm/r must be provided to infinitely separate them).​​

Electric…

Fields

  • Magnitude: E = F/q = kQ/r².

2.png3.png

Potential (Ve)

  • Definition: “The electrical potential at a point P is the work done per unit charge for a small positive test charge to be brought from infinity to that point”.

  • May be visualized as the height of a flat surface.

Potential energy (EP)

  • Definition: “The electrical potential energy at a point P is the work done for a small positive test charge to be brought from infinity to that point”.

  • Charge sign must always be taken into account.

Parallel plates

FieldLinesEquipotentials.png

  • Explanation: long oppositely charge plates.

  • Field is uniform in the region between the plates.

  • Edge effect: field becomes weaker at the edges.

​Equipotential surface

  • Explanation: consists of those points that have the same potential, i.e. which are at the same distance from the source (referred to as zero potential), and where masses or charges move without work being done on or by then.

  • Field lines are cut perpendicularly by the equipotential surfaces.

10.2 Fields at Work

Graphical interpretation of gravitational field strength and potential

  • Going upstream in the field (against) means going to a higher potential, so gain in the potential.

  • Going downstream in the field (in favour) means going to a lower, so loss in the potential.

  • Gradient of a graph of gravitational potential against distance is the gravitational field strength. g = -∆Vg/∆r

EarthMoon.pnggradient.png

 

Inside a planet​​

3.png

 

Weightlessness

Feeling weightless for an astronaut in orbit around the Earth is a consequence of both ship and the astronaut “falling freely”, with the same acceleration towards the center of planet, so that there is no normal force.

Orbits

  • Orbital speed (vorbit): sqrt(GM/r).

  • Orbital period (Torbit): sqrt[4π²r³/(GM)].

  • Polar orbit: for satellites close to the Earth’s surface (100 km).

  • Geostationary orbit: for geosynchronous satellites, whose period is equal to 24 hours.

Total energy (ET) = kinetic energy (EK) + gravitational potential energy (EP).

  • ET = EK + EP = 1/2mv² – GMm/r = GMm/2r – GMm/r = – GMm/2r

  • Graph of the kinetic, potential and total energy of a mass in circular orbit around a planet as function of distance.

  • Increase in the orbit: total energy increases, potential energy increases and kinetic energy decreases.

  • Air friction: radius decreases, causing the total energy to decrease, potential energy to decrease and kinetic energy to increase.

  • Launching a body from a planet’s surface cases:

    • If total energy is positive: object will follow a hyperbolic path and never return.

    • If total energy is zero: object will follow a parabolic path to infinity, where it will stop.

    • If total energy is negative: object will go into a circular or elliptical orbit or crash.

  • Escape velocity: “minimum speed of object to escape gravitational field of planet/travel to infinity, starting at the surface of a planet, without energy input”.

    • vescape: sqrt(2GM/r).​

Graphical interpretation of electric field strength and potential

FieldIntensity.png
  • Electric field strength is the force per unit charge, and thus, the area under the graph of electrical field strength against distance is the work per unit charge, i.e. the electric potential charge.

WorkGraph.png

Inside a hollow conducting charged sphere

  • As the sphere is a conductor, all the surplus must reside on the outside of the sphere.

    • Charges will move until they are as far apart as possible and in equilibrium equidistant on the surface.​

  • Inside a sphere, the force acting on a test charge are always equal in sizer and opposite in direction, and thus, cancel out: E = 0, which is the gradient ∆Ve/∆r, which means that V is constant.

7.gif

Charges moving in magnetic and electric fields

Magnetic fields: force will be at right angles to velocity and magnetic field strength.

  • Circular path: when the charge’s direction is perpendicular to magnetic field strength.

    • Magnetic force = centripetal force​

8.jpg
  • Helical path: charge’s movement when direction is not perpendicular to magnetic field strength.

9.png

Electric field produced by the uniform field in parallel plates

  • Only vertical acceleration, no horizontal.

  • Combination of magnetic and electric fields opposing each other, which may generate balance of forces and the charge may move in a horizontal path.

​Inverse square law behavior

  • Geometric explanation: influence per unit area reduces to the power of 2.

10.png

ELECTRIC FIELD

The space around an electric charge, where it exerts a force on another charge is an electric field.
Electric force, like the gravitational force acts between the bodies that are not in contact with each other. To understand these forces, we involve the concept of force field. When a mass is present somewhere, the properties of space in vicinity can be considered to be so altered in such a way that another mass brought to this region will experience a force there. The space where alteration is caused by a mass is called its Gravitational field and any other mass is thought of as interacting with the field and not directly with the mass responsible for it.

Similarly an electric charge produces an electric field around it so that it interacts with any other charges present there. One reason it is preferable not to think of two charges as exerting forces upon each other directly is that if one of them is changed in magnitude or position, the consequent change in the forces each experiences does not occur immediately but takes a definite time to be established. This delay cannot be understood on the basis of coulomb law but can be explained by assuming (using field concept) that changes in field travel with a finite speed. (≈ 3 × 108 m / sec).

Electric field can be represented by field lines or line of force.
The direction of the field at any point is taken as the direction of the force on a positive charge at the point.

Electric field intensity due to a charge q at any position () from that charge is defined as

where is the force experienced by a small positive test charge q0 due to charge q.
Its SI unit is NC–1. It is a vector quantity.
If there are more charges responsible for the field, then 
where are the electric field intensities due to charges q1, q2, q3…..respectively.

ELECTRIC LINES OF FORCE

These are the imaginary lines of force and the tangent at any point on the lines of force gives the direction of the electric field at that point.

PROPERTIES OF ELECTRIC LINES OF FORCE

  1. The lines of force diverge out from a positive charge and converge at a negative charge. i.e. the lines of force are always directed from higher to lower potential.

  1. The electric lines of force contract length wise indicating unlike charges attract each other and expand laterally indicating like charges repel each other.

  1. The number of lines that originate from or terminate on a charge is proportional to the magnitude of charge.

i.e., 

  1. Two electric lines of force never intersect each other.
  2. They begin from positive charge and end on negative charge i.e., they do not make closed loop (while magnetic field lines form closed loop).

  1. Where the electric lines of force are
  1. close together, the field is strong (see fig.1)
  2. far apart, the field is weak (see fig.2)

  1. Electric lines of force generate or terminate at charges /surfaces at right angles.

ELECTRIC FIELD FOR CONTINUOUS CHARGE DISTRIBUTION

If the charge distribution is continuous, then the electric field strength at any point may be calculated by dividing the charge into infinitesimal elements. If dq is the small element of charge within the charge distribution, then the electric field at point P at a distance r from charge element dq is

Non conducting sphere (dq is small charge element)

dq = λdl (line charge density)
= σ ds (surface charge density)
= ρdv (volume charge density)
The net field strength due to entire charge distribution is given by

where the integration extends over the entire charge distribution.

Note:- Electric field intensity due to a point charge q, at a distance (r1 + r2) where r1 is the thickness of medium of dielectric constant K1 and r2 is the thickness of medium of dielectric constant K2 as shown in fig. is given by

CALCULATION OF ELECTRIC FIELD INTENSITY FOR A DISTRIBUTION OF DIRECT AND CONTINUOUS CHARGE

  1. Fix origin of the coordinate system where electric field intensity is to be found.
  2. Draw the direction of electric field intensity due to the surrounding charges considering one charge at a time.
  3. Resolve the electric field intensity in x and y-axis respectively and find ΣEx and ΣEy
  4. The resultant intensity is and where θ is the angle between and x-axis.
  5. To find the force acting on the charge placed at the origin, the formula F = qE is used.

ENERGY DENSITY
Energy in unit volume of electric field is called energy density and is given by
,
where E = electric field and εo= permittivity of vacuum

ELECTRIC FIELD DUE TO VARIOUS CHARGE DISTRIBUTION

  • Electric Field due to an isolated point charge


  • A circular ring of radius R with uniformly distributed charge



When x >> R, 
[The charge on ring behaves as point charge]
E is max when . Also Emax

  • A circular disc of radius R with uniformly distributed charge with surface charge density σ


  • An infinite sheet of uniformly distributed charges with surface charge density σ


  • A finite length of charge with linear charge density


and 
Special case :
For Infinite length of charge, 
∴  and  

  • Due to a spherical shell of uniformly distributed charges with surface charge density σ


Ein = 0 (x < R)

  • Due to a solid non conducting sphere of uniformly distributed charges with charge density ρ


 
 

  • Due to a solid non-conducting cylinder with linear charge density λ


Eaxis = 0, ,
,
In above cases, 

KEEP IN MEMORY

  1. If the electric lines of force are parallel and equally spaced, the field is uniform.
  2. If E0 and E be the electric field intensity at a point due to a point charge or a charge distribution in vacuum and in a medium of dielectric constant K then

E = KE0

  1. If E and E’ be the electric field intensity at a point in the two media having dielectric constant K and K’ then

  1. The electric field intensity at a point due to a ring with uniform charge distribution doesn’t depend upon the radius of the ring if the distance between the point and the centre of the ring is much greater than the radius of the ring. The ring simply behaves as a point charge.
  2. The electric field intensity inside a hollow sphere is zero but has a finite value at the surface and outside it (; x being the distance of the point from the centre of the sphere).
  3. The electric field intensity at a point outside a hollow sphere (or spherical shell) does not depend upon the radius of the sphere. It just behaves as a point charge.
  4. The electric field intensity at the centre of a non-conducting solid sphere with uniform charge distribution is zero. At other points inside it, the electric field varies directly with the distance from the centre (i.e. E ∝ x; x being the distance of the point from the centre). On the surface, it is constant but varies inversely with the square of the distance from the centre (i.e.). Note that the field doesn’t depend on the radius of the sphere for a point outside it. It simply behaves as a point charge.
  5. The electric field intensity at a point on the axis of non-conducting solid cylinder is zero. It varies directly with the distance from the axis inside it (i.e. E ∝ x). On the surface, it is constant and varies inversely with the distance from the axis for a point outside it (i.e. ).

MOTION OF A CHARGED PARTICLE IN AN ELECTRIC FIELD

Let a charged particle of mass m and charge q be placed in a uniform electric field, then electric force on the charge particle is 
∴ acceleration,  (constant)

  • The velocity of the charged particle at time t is,

v = u + at = at =  (Particle initially at rest) or 

  • Distance travelled by particle is 
  • Kinetic energy gained by particle, 

If a charged particle is entering the electric field in perpendicular direction.

Let  and the particle enters the field with speed u along x-axis.
Acceleration along Y-axis, 
The initial component of velocity along y-axis is zero. Hence the deflection of the particle along y-axis after time t is ;

…… (i)
Distance covered by particle in x-axis,
x = ut …… (ii) ( acceleration ax = 0)
Eliminating t from equation (i) & (ii),
 i.e. y ∝ x2.
This shows that the path of charged particle in perpendicular field is parabola.
If the width of the region in which the electric field exists be l then

  1. the particle will leave the field at a distance from its original path in the direction of field, given by 
  2. The particle will leave the region in the direction of the tangent drawn to the parabola at the point of escape.
  3. The velocity of the particle at the point of escape is given b

  1. The direction of the particle in which it leaves the field is given by


ELECTRIC DIPOLE

Two equal and opposite charges separated by a finite distance constitute an electric dipole. If –q and +q are charges at distance 2l apart, then dipole moment,


Its SI unit is coulomb metre.
Its direction is from –q to +q. It is a vector quantity.
The torque τ on a dipole in uniform electric field as shown in figure is given by, 
So τ is maximum, when dipole is ⊥ to field & minimum (=0) when dipole is parallel or antiparallel to field.
If  and 
Then  

The work done in rotating the dipole from equilibrium through an angle dθ is given by

and from θ1 → θ2, 

If θ1 = 0 i.e., equilibrium position, then

Work done in rotating an electric dipole in uniform electric field from θ1 to θ2 is W = pE (cosθ1 – cosθ2)

Potential energy of an electric dipole in an electric field is,
 i.e. U = –pE cosθ
where θ is the angle betweenand .
We can also write 

ELECTRIC FIELD DUE TO AN ELECTRIC DIPOLE

  • Along the axial line (or end-on position)


and are parallel
 when x >> l

  • Along equatorial line (or broadside on position)


 when x >>l
When  and are anti parallel then,
Eax = 2 Eeq

  • At any point (from the dipole)



Electric field intensity due to a point charge varies inversely as cube of the distance and in case of quadrupole it varies inversely as the fourth power of distance from the quadrupole.

ELECTRIC FORCE BETWEEN TWO DIPOLES

The electrostatic force between two dipoles of dipole moments p1 and p2 lying at a separation r is
 when dipoles are placed coaxially
 when dipoles are placed perpendicular to each other.

KEEP IN MEMORY

  1. The dipole moment of a dipole has a direction from the negative charge to the positive charge.
  2. If the separation between the charges of the dipole is increased (or decreased) K-times, the dipole moment increases (or decreases) by K-times.
  3. The torque experienced by a dipole placed in a uniform electric field has value always lying between zero and pE, where p is the dipole moment and E, the uniform electric field. It varies directly with the separation between the charges of the dipole.
  4. The work done in rotating a dipole in a uniform electric field varies from zero (minimum) to 2pE (maximum). Also, it varies directly with the separation between the charges of the dipole.
  5. The potential energy of the dipole in a uniform electric field always lies between +pE and –pE.
  6. The electric field intensity at a point due to an electric dipole varies inversely with the cube of the distance of the point from its centre if the distance is much greater than the length of the dipole.
  7. The electric field at a point due to a small dipole in end-on position is double of its value in broad side-on position,

i.e. EEnd-on = 2EBroad side-on

  1. For a small dipole, the electric field tends from infinity at a point very close to the axis of the dipole to zero at a point at infinity.
  2. The force between two dipoles increases (or decreases) by K4-times as the distance between them decreases (or increases) by K-times.
  3. Time period of a dipole in uniform electric field is


where I = moment of inertia of the dipole about the axis of rotation.

MOVING CHARGES AND MAGNETISM OERSTED EXPERIMENT In 1802 Gian Domenico Romagnosi , an Italian lawyer and judge, found that a steady …
MOVING CHARGES AND MAGNETISM

OERSTED EXPERIMENT

In 1802 Gian Domenico Romagnosi, an Italian lawyer and judge, found that a steady electric current flowing in wire affected a magnetic needle placed near it. He published his observation in a local newspaper (called Gazzetta di Trentino). But nobody noticed this phenomenon.

 

In 1820 Hans Christian Oersted (a Danish Physicist) rediscovered this phenomenon. He noted that a magnetic compass needle, brought close to a straight wire carrying a steady electric current, aligned itself perpendicular to the wire i.e., the direction of magnetic field is tangential to a circle which has the wire as centre, and which has its plane perpendicular to the wire (Fig 1-a).

 

Oersted also noticed that on reversing the direction of current; the direction of magnetic field is reversed.
In first case when current is in upward direction, magnetic field is clockwise (Fig 1-a) and when the current is downward, magnetic field is anticlockwise (Fig. 1-b).

MAGNETIC FIELD

It is a region of space around a magnet or current carrying conductor or a moving charge in which its magnetic effect can be felt.
The conductor carrying current is electrically neutral but a magnetic field is associated with it.
The SI unit of magnetic field induction is tesla (T) or weber/m2 and cgs unit is gauss. 1 gauss = 10–4 T
Comparison between electric field and magnetic field

MAGNETIC FIELD DUE TO CURRENT CARRYING CONDUCTOR, BIOT-SAVART’S LAW

The magnetic induction at any point outside the current path due to a small current element of length (in the direction of the current) is given by Biot-Savart’s law
Also, where  is the drift velocity of charge
where μ0 = 4π × 10–7 TmA–1.

 

Direction of
The direction of is  perpendicular to both and , governed by the right hand thumb rule of the cross-product of   and .
The magnetic fields going into the page and coming out of the page are represented as follows :

MAGNETIC FIELD DUE TO VARIOUS CURRENT CARRYING CONDUCTORS

MAGNETIC FIELD DUE TO FINITE SIZED CONDUCTOR

 

ELUCIDATION
r = a sec θ,    = a tanθ
(Pointing into the plane of paper)

MAGNETIC FIELD NEAR THE END OF A FINITE SIZED CONDUCTOR

MAGNETIC FIELD DUE TO AN INFINITELY LONG CONDUCTOR

 

ELUCIDATION
Magnetic field in the case of infinitely long wire

MAGNETIC FIELD NEAR THE END OF A LONG CONDUCTOR

 

ELUCIDATION
r = a cosec θ, l = a cot θ,
dl = a cosec2θ dθ
      

MAGNETIC FIELD DUE TO A CURRENT CARRYING COIL

MAGNETIC FIELD AT A POINT ON THE AXIS OF SYMMETRY OF A CIRCULAR COIL, AT A DISTANCE “X” FROM ITS CENTRE
B = μ0NI a2/2(a2 + x2)3/2
or,   
N =  total number of turns
a = coil radius
The direction of  is given by Right hand screw rule.

 

Right hand screw rule : If direction of rotation of right handed screw-head is the direction of current in a circular conductor then the direction of its advance is the direction of magnetic field. This is applicable even if the current, magnetic field are interchanged, as in case of current flowing through a straight conductor.

 

ELUCIDATION
Let for a particular angle, position of small length element dl is given by its coordinates as
Now, ,
Also we have .
Now   at any instant
,

 

If number of turns of coil are N, then

 

MAGNETIC FIELD AT THE CENTRE OF A CIRCULAR COIL
B = μ0NI / 2a

 

MAGNETIC FIELD AT THE CENTRE OF A CIRCULAR ARC CARRYING CURRENT
where θ is in radian.
In this case the direction of magnetic field is into the page.

MAGNETIC FIELD INSIDE A CURRENT CARRYING SOLENOID

  • Finite size solenoid
  • Near the end of a finite solenoid
  • In the middle of a very long solenoid, B = μ0 n I
  • Near the end of a very long solenoid
n is the number of turns per unit length of solenoid.
  • Magnetic field in the endless solenoid (toroid) is same throughout and is μ0nI.
  • Magnetic field outside a solenoid or toroid is zero.

AMPERE’S CIRCUITAL LAW

The line integral of magnetic field across a closed loop is equal to 40 times the net correct inside the loop
i.e.
where I  is the net current inside the loop.
  • The direction of the magnetic field at a point on one side of a conductor of any shape is equal in magnitude but opposite in direction of the field at an equidistant point on the other side of the conductor.
  • If the magnetic field at a point due to a conductor of any shape is Bo if it is placed in vacuum then the magnetic field at the same point in a medium of relative permeability  μr  is given by .
  • If the distance between the point and an infinitely long conductor is decreased (or increased) by K-times then the magnetic field at the point increases (or decreases) by K-times.
  • The magnetic field at the centre of a circular coil of radius smaller than other similar coil with greater radius is more than that of the latter.
  • For two circular coils of radii R1 and R2 having same current and same number of turns,where B1 and B2 are the magnetic fields at their centres.
  • The magnetic field at a point outside a thick straight wire carrying current is inversely proportional to the distance but magnetic field at a point inside the wire is directly proportional to the distance.

 

KEEP IN MEMORY
  1. If in a coil the current is clockwise, it acts as a South-pole. If the current is anticlockwise, it acts as North-pole.
  1. No magnetic field occurs at point P, Q and R due to a thin current element .
  1. Magnetic field intensity in a thick current carrying conductor at any point x is
  1. Graph of magnetic field B versus x
  1. Magnetic field is zero at all points inside a current carrying hollow conductor.

MAGNITUDE AND DIRECTION OF MAGNETIC FIELD DUE TO DIFFERENT CONFIGURATION OF CURRENT CARRYING CONDUCTOR

FORCE ON A CONDUCTOR

The force on a conductor is given by
F = BIl sin α
where,
l is the length of the conductor in meter;
B is the flux density of field in tesla (Wb/m2);
I is the current in ampere and
α is the angle which the conductor makes with the direction of the field.
Special case :
If α = 90°, then  F = BIl
The direction of the force is given by Fleming’s left hand rule.

TORQUE ON A COIL

The torque acting on a rectangular coil placed with its plane parallel to a uniform magnetic field of flux density B is given by
τ = BINA
where N is the number of turns in the coil, A is the area and I is the current.
If the plane of the coil makes an angle α with the direction of the field, then
τ = BINA cos α

FORCE ACTING ON A CHARGED PARTICLE MOVING IN A UNIFORM MAGNETIC FIELD

The force acting on a particle having a charge q and moving with velocity  in a uniform magnetic field  is given by  
,
where θ is the angle between and

 

Case (i)
If θ = 0, F = 0. Also if θ = 180°, F = 0
If a charged particle enters a uniform magnetic field in the direction of magnetic field or in the opposite direction of magnetic field, the force acting on the charged particle is zero.

 

Case (ii)
If θ = 90°, F = qvB
In this case the force acting on the particle is maximum and this force acts as centripetal force which makes the charged particle move in a circular path.
where r is the radius of the circular path.
It is important to note that this force cannot change the speed of the charged particle and hence its kinetic energy. But it changes the velocity of charged particle (due to change in direction) and hence also causes a change in momentum. Also the work done by the force is zero as the force is acting perpendicular to the direction of motion.

 

Case (iii)
If θ = α is any other angle then the path taken is helical. The velocity of the charged particle can be split into two parts for better understanding.
v cos α : The force on charged particle due to this component is zero. This component is responsible in moving the charged particle uniformly in the direction of .
v sin α : The force acting on charged particle due to this component is
This acts as the centripetal force and moves the particle in a circular path.

 

The combined effect of these two is a helical path.
A charged particle entering a uniform magnetic field at an angle executes helical path.
Radius of the helix,
Angular frequency of rotation,
Pitch of the helix =

 

Direction of force  :We can use the rule of cross product. The direction of is perpendicular to the plane containing   and  and can be found by right hand thumb rule. It is important to note that if q is positive, we will get the correct direction of  by right hand thumb rule. But if q is negative, we have to reverse the direction of force.

 

FLEMING’S LEFT HAND RULE
It states that if the fore finger, the central finger and the thumb of the left hand are stretched at right angles to each other then if the central finger represents the direction of current and fore finger represents field, the thumb will represent the direction of motion or force experienced by the current carrying conductor.

FORCE BETWEEN TWO PARALLEL CURRENTS

When a current flows in a conductor, the free charges (electrons in case of a metal wire) move. Each free charge movement generates a force which adds up to give the force on the conductor.

 

Force between infinitely long conductors placed parallel to each other at distance d.
Force per unit length =
where l is the length of wire.
If currents are pointing in same direction, the force is of attractive nature and if currents are oppositely directed the force is of repulsive nature.

 

LORENTZ FORCE EQUATION
For a charged particle q moving in a region of simultaneously applied electric field   and magnetic field , the force experienced by it is given by
Torque on a current loop in uniform magnetic field is given by  where  is the magnetic moment of coil.
 where is the unit vector normal to the plane of the loop.

 

KEEP IN MEMORY
  1. No force acts on a charged particle if it enters a magnetic field in a direction parallel or antiparallel to the field.
  2. A finite force acts on a charged particle if it enters a uniform magnetic field in a direction with finite angle with the field.
  3. If two charged particles of masses m1and m2 and charges q1 and q2 are projected in a uniform magnetic field with same constant velocity in a direction perpendicular to the field then the ratio of their radii (R1: R2) is given by
  1. The force on a conductor carrying current in a magnetic field is directly proportional to the current, the length of conductor and the magnetic field.
  2. If the distance between the two parallel conductors is decreased (or increased) by k-times then the force between them increases (or decreases) k-times.
  3. The momentum of the charged particle moving along the direction of magnetic field does not change, since the force acting on it due to magnetic field is zero.
  4. Lorentz force between two charges q1 and q2 moving with velocity v1, v2 separated by distance r is given by
  1. If the charges move, the electric as well as magnetic fields are produced. In case the charges move with speeds comparable to the speed of light, magnetic and electric force between them would become comparable.
  2. A current carrying coil is in stable equilibrium if the magnetic dipole moment , is parallel to   and is in unstable equilibrium when is antiparallel to .
  3. Magnetic moment is independent of the shape of the loop. It depends on the area of the loop.
  4. A straight conductor and a conductor of any shape in the same plane and between the same two end points carrying equal current in the same direction, when placed in the same magnetic field experience the same force.
  5. There is net repulsion between two similar charges moving parallel to each other in spite of attractive magnetic force between them. This is because of electric force of repulsion which is much more stronger than the magnetic force.
  6. The speed of the charged particle can only be changed by an electric force.

MOVING COIL GALVANOMETER

The moving coil galvanometer was first devised by Kelvin and later on modified by D’Arsonval. This is used for detection and measurement of small electric current.

 

PRINCIPLE
The principle of a moving coil galvanometer is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque.

 

CONSTRUCTION
A moving coil ballistic galvanometer is shown in figure.
It essentially consists of a rectangular coil PQRS or a cylindrical coil of large number of turns of fine insulated wire wound over a non-conducting frame of ivory or bamboo. This coil is suspended by means of phosphor bronze wire between the pole pieces of a powerful horseshoe magnet NS. The poles of the magnet are curved to make the field radial. The lower end of the coil, is attached to a spring of phosphor-bronze wire. The spring and the free ends of phosphor bronze wire are joined to two terminals T2 and T1 respectively on the top of the case of the instrument. L is a soft iron core. A small mirror M is attached on the suspension wire. Using lamp and scale arrangement, the deflection of the coil can be recorded. The whole arrangement is enclosed in a non-metallic case.

 

THEORY
Let the coil be suspended freely in the magnetic field.
Suppose, n = number of turns in the coil
A = area of the  coil
B = magnetic field induction of radial magnetic field in which the coil is suspended.
Here, the magnetic field is radial, i.e.,  the plane of the coil always remains parallel to the direction of magnetic field, and hence the torque acting on the coil
τ = niAB … (1)

 

Due to this torque, the coil rotates. As a result, the suspension wire gets twisted. Now a restoring torque is developed in the suspension wire. The coil will rotate till the deflecting torque acting on the coil due to flow of current through it is balanced by the restoring torque developed in the suspension wire due to twisting. Let C be the restoring couple for unit twist in the suspension wire and θ be the angle through which the coil has turned. The couple for this twist θ is Cθ.
In equilibrium, deflecting couple = restoring couple
∴   ni AB = Cθ  or i = Cθ/ (nAB)
or i = Kθ (where C/nAB = K) … (2)
K is a constant for galvanometer and is known as galvanometer constant.
Hence
Therefore, the deflection produced in the galvanometer is directly proportional to the current flowing through it.

 

Current sensitivity of the galvanometer : The current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current is passed through it.
We know that, niAB = Cθ
Current sensitivity, is =
where C = restoring couple per unit twist
The SI unit of current sensitivity is radian per ampere or deflection per ampere.

 

Voltage sensitivity of the galvanometer : The voltage sensitivity of the galvanometer is defined as the deflection produced in the galvanometer when a unit voltage is applied  across the terminals of the galvanometer.
Voltage sensitivity,  
If R be the resistance of the galvanometer and a current is passed through it, then V = iR
Voltage sensitivity, Vs =
The SI. unit of voltage sensitivity is radian per ampere or deflection per ampere.

 

Conditions for sensitivity : A galvanometer is said to be more sensitive if it shows a large deflection even for a small value of current.
We know that,
For a given value of i, θ will be large if (i) n is large, (ii) A is large, (iii) B is large, and (iv) C is small.
Regarding above factors, n and A cannot be increased beyond a certain limit. By increasing n, the  resistance of the galvanometer will increase and by increasing A, the size of the galvanometer will increase. So, the sensitivity will decrease. Therefore, B is increased. The value of B can be increased by using strong horseshoe magnet. Further, the value of C can be decreased. The value of C for quartz and phosphor-bronze is very small. So, the suspension wire of quartz or phosphor-bronze is used. The value of C is further decreased if the wire is hammered into a flat strip.

HALL EFFECT

When a current passes through a slab of material in the presence of a transverse magnetic field, a small potential difference is established in a direction perpendicular to both, the current flow and the magnetic field. This effect is called Hall effect.

 

The voltage thus developed is called Hall voltage.
Hall effect enables us to :
  • Determine the sign of charge carriers inside the conductor.
  • Calculate the number of charge carriers per unit volume.

 

EXPLANATION
Let us consider a conductor carrying current in +X direction. The magnetic field is applied along +Y direction. Consider two points P1 and P2 on the conductor and connect a voltmeter between these points. If no magnetic field is applied across the conductor, then the points P1 and P2 will be at same potential and there will be no deflection in the galvanometer. However, if a magnetic field is applied as shown in the figure, then the Lorentz force acts on electrons as shown in the figure.
The Lorentz force on electrons Fm = –e acts in the downwards direction.
Now there may be two cases:
Case I : If the charge particles are negatively charged, then these negative charges will accumulate at the point P2 and therefore P2 will be at lower potential than P1.
Case II : If the charged particles are positively charged, then the point P2 will be at higher potential than P1.

 

Magnitude of hall voltage :
Let w be width and A be the cross-sectional area of the conductor. If e is magnitude of charge or the current carrier (electron or hole).
The force on the current carrier due to magnetic field  B,
Fm = Bevd
Here, vd is drift velocity of the current carries.
Due to the force Fm, the opposite charges build up at the points P1 and P2 of the conductor.
If VH is Hall voltage developed across the two faces, then the strength of electric field due to Hall voltage is given by
EH = .  
Here   w = P1P2.
This electric field exerts an electric force on the current carries in a direction opposite to that of magnetic force.
The magnitude of this force is
In equilibrium condition,   Fe = Fm,
or
Now, drift velocity of current carrier is given by,     vd =
where n is the number of current carries per unit volume of the strip.
Hall resistance ,
Hall voltage

 

KEEP IN MEMORY
Hall effect can determine nature of current (charge) carriers in the material. i.e. whether the charge is +ve or –ve.
Hall voltage
where n is the density of charge carriers
b = thickness of plate,
B = magnetic field,
I  = current flowing through plate,
A = area of cross-section of plates

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