It is the product of strength of pole and magnetic length of the magnet. M = m (2 l)
Its SI unit is Ampere m2
Bar Magnet as an Equivalent Solenoid
Let a be radius of solenoid
i be strength of current passed through the solenoid
n be number of turns per unit length of solenoid
2 l be length of solenoid with centre O
We have to calculate magnetic field at point P on the axis of solenoid such that OP = r.
Consider a small element of width dx of a solenoid at a distance x from O
This is the required expression for magnetic field on the axial line of a short bar magnet. Hence, a finite solenoid carrying current is equivalent to bar magnet.
Electric Dipole Moment of a Revolving Electron or Bohr Magneton
Electrons revolves around nucleus in an atom. This revolving electron is like a loop of current with a definite dipole moment. If electron revolves in anticlockwise direction then electric current will flow in clockwise direction. Thus upper face of the electron loop behaves as south pole whereas lower face acts as north pole.
Let e be the charge on revolving electron and r be the radius of orbit
ω be the uniform angular velocity
Electric Current (i) = Charge (e) / Time (T)
Here T is the period of revolution of electron T = 2π / ω
i = e / (2π / ω) = eω / 2π
i = eω / 2π
Area of orbit in which electron is revolving, A = πr2
Magnetic moment of an atom is given as M = i A = (eω / 2π) ( πr2 )
M = 1/2 eωr2 ……(A)
According to Bohr’s postulate; m v r = nh / 2 π
m (ωr) r = n h / 2 π
ωr2 = nh / 2 πm
Now Substituting the value of ωr2 in equation (A), we get
M = (1/2) (e) (nh / 2 πm )
M = nhe / 4 πm
This is the required expression for orbital magnetic moment of an electron revolving in nth orbit.
It is defined as minimum magnetic dipole moment of an orbit due to motion of an electron in first stationary orbit (n = 1) of the atom.
e = 1.6 x 10-19 C
h = 6.63 x 10-34 Js
m = 9.1 x 10-31 kg
μB = eh / 4 πm = 1.6 x 10-19 x 6.63 x 10-34 / 4 x 3.14 x 9.1 x 10-31
μB = 9.27 x 10-24 Ampere metre2
Gyromagnetic Ratio of Electron
It is the ratio of magnetic moment to the angular moment of an electron
Gyromagnetic Ratio = Magnetic Moment / Angular Moment = (1/2 eωr2 ) / (mωr2 )
Gyromagnetic Ratio = e / 2m = 8.8 x 1010 C / kg
Question: What are the proof of earth’s magnetism?
Solution: Earth is itself a huge magnet because of the following evidences:
When a magnet is suspended freely with a thread and is allowed to rotate in horizontal plane it comes to rest along the north south direction.
When a piece of soft iron is buried under the earth, it is found to attain the properties of a magnet after sometime.
On plotting the magnetic field lines of a magnet, neutral points are obtained.
Concept: Earth’s crust has molten form of iron which contain charged ions. Earth rotates from west to east direction. Thus these charged ions produces a circulating current. The circulating current produced is mainly responsible for magnetic field of earth.
Some terms related to Earth’s Magnetism
Geographic Axis: The straight line passing through the geographic north and south poles of earth is defined as the geographic axis.
Magnetic Axis:The straight line passing through magnetic north and south poles of earth is commonly known as magnetic axis.
Geographic Meridian: The vertical plane passing through the geographic north and south poles is known as geographic meridian.
Magnetic Meridian: The vertical passing through the magnetic axis of freely suspended magnet is defined as Magnetic meridian.
Magnetic Elements of Earth
Magnetic elements of earth at any point are the quantities which gives the complete description about the magnitude and direction of magnetic field at that point.
There are three magnetic elements of earth as discussed below:
Magnetic Declination (θ)
Angle of dip or Magnetic inclination (δ)
Horizontal component (H) of earth’s magnetic field
Magnetic Declination (θ)
The angle between the geographic axis and magnetic axis at a place is defined as the magnetic declination at that point.
It is also defined as angle between magnetic meridian and geographic meridian at that place.
Angle of Dip or Magnetic Inclination (δ)
Magnetic dip or magnetic inclination at a place is defined as the angle which net earth’s magnetic field makes with horizontal in magnetic meridian.
Angle of dip is measured by a device commonly known as dip circle .
At equator, total magnetic field is parallel to horizontal, i.e., δ = 0
At earth’s magnetic polesδ= 90
Horizontal Component of Earth’s Magnetic Field
It is the component of net Earth’s magnetic field in horizontal direction. It is represented by H.
Earth’s magnetic field has a horizontal component only at magnetic equator.
At any point magnetic field can be resolved as horizontal component and vertical component
BH = Bcos δ BV = Bsinδ
Net Magnetic field, B = √ (BH)2 + (BV)2
tan δ = BV / BH
Instrument used to measure horizontal component of earth’s magnetic field is commonly known as vibration magnetometer and a deflection magnetometer.
1. At Equator δ = 0°
BV = B sin 0° = 0
BH = B cos 0° = B
At equator vertical component of earth’s magnetic field is zero and only horizontal component exists.
2. At Pole δ = 90°
BV = B sin90° = B
BH = B cos 90° = 0
At a place on the magnetic pole, Horizontal component of earth’s magnetic field is zero and only vertical component exists.
Question : At what angle horizontal and vertical component of earth’s magnetic field have equal value?
Solution: BH = BV
B cosδ = B sinδ
tanδ = 1
δ = 45°
Question : Why does a compass become in effective in submarine?
Solution: The body of submarine is made of iron when a compass is kept in a submarine it experiences force from all side and thus the effective force on submarine becomes zero. Hence a compass becomes in effective in a submarine.
Question : Angle of dip at a place is 30°. Vertical component of Earth’s magnetic field at that point is 0.2 G. Calculate total magnetic field at that point.
Solution: BV = B sin30° 0.2 = B (1/2) B = 0.4 G B = 0.4 × 10-4 Tesla
Question : What will be the measure of vertical component of Earth’s Magnetic field at a point where dip is 60° and the horizontal component of earth’s magnetic field is 0.2 x 10-4 T ?
Solution: tan 60° = BV / BH
BV = √3 x 0.2 x 10-4
BV = 0.2√3 x 10-4 Tesla
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