Moving coil galvanometer is an electromagnetic device that can measure small values of current. It is also known as Weston galvanometer. It works on the principle that when a current loop is placed in an external magnetic field, it experiences torque, and the value of torque can be changed by changing the current in the loop
Moving coil galvanometer consists of permanent horse-shoe magnets, coil, soft iron core, pivoted spring, non-metallic frame, scale and pointer.
We know that a current loop having N number of turns, and the cross sectional area A, carrying current i, when placed in and along the direction of external magnetic field B, experiences a torque given by: ԏ = NiAB
The pivoted spring of spring constant k would oppose the above torque with restoring torque C given by: C = kΦ
Here, Φ is the angular deflection of spring
Both, the torque, and the restoring torque would be equal:
kΦ = NiAB
Φ = NiAB/k
In the above equation, except for current, every other quantity on the right hand side is constant for a galvanometer, hence:
Φ ∝ i
So, the angular deflection Φ produced in the pointer could be measured in terms of current in the scale calibrated on the basis of above equations. To use galvanometer as an ammeter (to measure higher values of current), we need to connect a shunt wire, with very small resistance(Rs), in parallel with the galvanometer (which have very low resistance of Rg
Equivalent resistance R of ammeter will be:
R = RgRs/(Rg + Rs)
Rg ˃˃ Rs
∴ R = RgRs/Rg = Rs
So, the equivalent resistance of ammeter is very less, which is a must for sensitivity of ammeter to be higher. Also, most of the current will pass through the shunt, thus protecting the galvanometer from any damage. Ammeter is connected in series with the circuit where current is to be measured
Current sensitivity(deflection per unit current) of galvanometer is given by:
Φ/i = NAB/k
To use galvanometer as a voltmeter, we need to connect a wire, with very high resistance(Rw˃˃Rg), in series with the galvanometer to ensure that our voltmeter equivalent resistance is high and so that it will draw a very small current. Equivalent resistance will be given by:
R = Rg + Rw = Rw
Voltmeter is connected in parallel with the circuit where voltage is to be measured.
Voltage sensitivity (deflection per unit voltage) of galvanometer is given by:
Φ/V = NAB/(kR)
Conversion of a Galvanometer into Ammeter and Voltmeter
Conversion of a Galvanometer into an Ammeter:
An ammeter is an instrument used to measure the current passing through a circuit. A galvanometer can be converted into ammeter by connecting a low resistance called shunt in parallel to the galvanometer.
Let G be the resistance of the galvanometer, I be the maximum current to be measured by the ammeter, and Ig be the maximum current that can be passed through the galvanometer. Then, (I-Ig) is the current passing through the shunt, S.
Since G and S are in parallel combination,
\(\begin{align} & p.d.\text{ }across\text{ }S=p.d.\text{ }across\text{ }G \\ & (IIg)S=IgG \\ & S=\frac{{{I}_{g}}}{I-{{I}_{g}}}\times G \\ \end{align}\)
This is the value of shunt which is connected in parallel to the galvanometer and this connection is converted into ammeter of range (0 – I) amperes. The effective resistance RA of the ammeter is given by
\(\begin{align} & \frac{I}{{{R}_{A}}}=\frac{1}{G}+\frac{1}{S} \\ & {{R}_{A}}=\frac{GS}{G+S} \\ \end{align}\)
So, the resistance of ammeter is smaller than S. Since S is low resistance, resistance of ammeter RA is very low and when it is connected in series in the circuits, it will not affect the current passing through the circuit.
Conversion of a Galvanometer into a Voltmeter:
A voltmeter is a device used to measure the potential difference between two points in a circuit. A galvanometer can be converted into voltmeter by connecting a high resistance called multiplier in series to the galvanometer.
Let G be the resistance of the galvanometer and Ig, the maximum deflection in the galvanometer. To measure the maximum voltage, V by the voltmeter, the high resistance R is connected in series. So,
\(\begin{align} & V={{I}_{g}}(R+G) \\ & or,{{I}_{g}}R=V-{{I}_{g}}G \\ & or,R=\frac{V}{{{I}_{g}}}-G \\ \end{align}\)
This equation gives the value of resistance R, which connected in series to the galvanometer, is converted into a voltmeter of range 0 – V volts. The effective resistance of the voltmeter Rv = R + G. since R is high, the resistance of the voltmeter Rv is high and it will not draw much current from the circuit.
1. Statement 1: The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as core
inside the coil.
Statement: Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
Solution:
The sensitivity of a moving-coil galvanometer is defined as angular deflection θ per unit current i. It is given as θ/i=nAB/k, where n is number of turns, A is coil area, B is magnetic field, and k is torsional constant of suspension wire. To increase sensitivity, B is increased by placing a soft iron core inside the coil. The soft iron has a high magnetic permeability and can be easily magnetized or demagnetized.
2. Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant?
Solution:
Low torsional constant facilitates greater deflection (allpha) in coil for given value of current and hence, sensitivity of galvanometer increases.
1. A moving coil galvanometer of resistance 100Ω is used as an ammeter using a resistance 0.1Ω. The maximum deflection current in the galvanometer is 100μA. Find the current in the circuit, so that the ammeter shows maximum deflection.
Solution:
A galvanometer of resistance G is converted to an ammeter by connecting a small shunt resistance S in parallel (see figure). Kirchhoff's loop law gives,
\({{i}_{g}}G-(i-{{i}_{g}})S=0,\Rightarrow i={{i}_{g}}(G+S)/S\)
The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter. Substitute the values to get
\(i={{i}_{g}}(G+S)/S=(100\times {{10}^{-6}})((100+0.1)/0.1))=100.1mA.\)
1. 2 moving coil galvanometers M1 and M2(having the same spring constants), have
the following specifications:
R1 = 10Ω, N1 = 30, A1 = 3.6×10-3m2, B1 = 0.25T
R2 = 14Ω, N2 = 42, A2 = 1.8×10-3m2, B2 = 0.5T
Find the ratios of current sensitivities between the 2 galvanometers.
Solution:
current sensitivity is given by:
Φ/i = NAB/k
(Φ/i)1 = (30×3.6×10-3×0.25)/k
(Φ/i)2 = (42×1.8×10-3×0.5)/k
∴(Φ/i)1/(Φ/i)2 = (30×3.6×10-3×0.25)/(42×1.8×10-3×0.5) = 5/7
2. 2 moving coil galvanometers M1 and M2(having the same spring constants),
have the following specifications:
R1 = 10Ω, N1 = 30, A1 = 3.6×10-3m2, B1 = 0.25T
R2 = 14Ω, N2 = 42, A2 = 1.8×10-3m2, B2 = 0.5T
Find the ratios of Voltage sensitivities between the 2 galvanometers.
Solution:
Φ/V = NAB/(KR)
(Φ/V)1 = (30×3.6×10-3×0.25)/(k×10)
(Φ/V)2 = (42×1.8×10-3×0.5)/(k×14)
∴ (Φ/V)1/(Φ/V)2 = (30×3.6×10-3×0.25×14)/(42×1.8×10-3×0.5×10) = 1
Moving coil galvanometer is an electromagnetic device that can measure small values of current. It is also known as Weston galvanometer. It works on the principle that when a current loop is placed in an external magnetic field, it experiences torque, and the value of torque can be changed by changing the current in the loop
Moving coil galvanometer consists of permanent horse-shoe magnets, coil, soft iron core, pivoted spring, non-metallic frame, scale and pointer.
We know that a current loop having N number of turns, and the cross sectional area A, carrying current i, when placed in and along the direction of external magnetic field B, experiences a torque given by: ԏ = NiAB
The pivoted spring of spring constant k would oppose the above torque with restoring torque C given by: C = kΦ
Here, Φ is the angular deflection of spring
Both, the torque, and the restoring torque would be equal:
kΦ = NiAB
Φ = NiAB/k
In the above equation, except for current, every other quantity on the right hand side is constant for a galvanometer, hence:
Φ ∝ i
So, the angular deflection Φ produced in the pointer could be measured in terms of current in the scale calibrated on the basis of above equations. To use galvanometer as an ammeter (to measure higher values of current), we need to connect a shunt wire, with very small resistance(Rs), in parallel with the galvanometer (which have very low resistance of Rg
Equivalent resistance R of ammeter will be:
R = RgRs/(Rg + Rs)
Rg ˃˃ Rs
∴ R = RgRs/Rg = Rs
So, the equivalent resistance of ammeter is very less, which is a must for sensitivity of ammeter to be higher. Also, most of the current will pass through the shunt, thus protecting the galvanometer from any damage. Ammeter is connected in series with the circuit where current is to be measured
Current sensitivity(deflection per unit current) of galvanometer is given by:
Φ/i = NAB/k
To use galvanometer as a voltmeter, we need to connect a wire, with very high resistance(Rw˃˃Rg), in series with the galvanometer to ensure that our voltmeter equivalent resistance is high and so that it will draw a very small current. Equivalent resistance will be given by:
R = Rg + Rw = Rw
Voltmeter is connected in parallel with the circuit where voltage is to be measured.
Voltage sensitivity (deflection per unit voltage) of galvanometer is given by:
Φ/V = NAB/(kR)
Conversion of a Galvanometer into Ammeter and Voltmeter
Conversion of a Galvanometer into an Ammeter:
An ammeter is an instrument used to measure the current passing through a circuit. A galvanometer can be converted into ammeter by connecting a low resistance called shunt in parallel to the galvanometer.
Let G be the resistance of the galvanometer, I be the maximum current to be measured by the ammeter, and Ig be the maximum current that can be passed through the galvanometer. Then, (I-Ig) is the current passing through the shunt, S.
Since G and S are in parallel combination,
\(\begin{align} & p.d.\text{ }across\text{ }S=p.d.\text{ }across\text{ }G \\ & (IIg)S=IgG \\ & S=\frac{{{I}_{g}}}{I-{{I}_{g}}}\times G \\ \end{align}\)
This is the value of shunt which is connected in parallel to the galvanometer and this connection is converted into ammeter of range (0 – I) amperes. The effective resistance RA of the ammeter is given by
\(\begin{align} & \frac{I}{{{R}_{A}}}=\frac{1}{G}+\frac{1}{S} \\ & {{R}_{A}}=\frac{GS}{G+S} \\ \end{align}\)
So, the resistance of ammeter is smaller than S. Since S is low resistance, resistance of ammeter RA is very low and when it is connected in series in the circuits, it will not affect the current passing through the circuit.
Conversion of a Galvanometer into a Voltmeter:
A voltmeter is a device used to measure the potential difference between two points in a circuit. A galvanometer can be converted into voltmeter by connecting a high resistance called multiplier in series to the galvanometer.
Let G be the resistance of the galvanometer and Ig, the maximum deflection in the galvanometer. To measure the maximum voltage, V by the voltmeter, the high resistance R is connected in series. So,
\(\begin{align} & V={{I}_{g}}(R+G) \\ & or,{{I}_{g}}R=V-{{I}_{g}}G \\ & or,R=\frac{V}{{{I}_{g}}}-G \\ \end{align}\)
This equation gives the value of resistance R, which connected in series to the galvanometer, is converted into a voltmeter of range 0 – V volts. The effective resistance of the voltmeter Rv = R + G. since R is high, the resistance of the voltmeter Rv is high and it will not draw much current from the circuit.
1. Statement 1: The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as core
inside the coil.
Statement: Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
Solution:
The sensitivity of a moving-coil galvanometer is defined as angular deflection θ per unit current i. It is given as θ/i=nAB/k, where n is number of turns, A is coil area, B is magnetic field, and k is torsional constant of suspension wire. To increase sensitivity, B is increased by placing a soft iron core inside the coil. The soft iron has a high magnetic permeability and can be easily magnetized or demagnetized.
2. Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant?
Solution:
Low torsional constant facilitates greater deflection (allpha) in coil for given value of current and hence, sensitivity of galvanometer increases.
1. A moving coil galvanometer of resistance 100Ω is used as an ammeter using a resistance 0.1Ω. The maximum deflection current in the galvanometer is 100μA. Find the current in the circuit, so that the ammeter shows maximum deflection.
Solution:
A galvanometer of resistance G is converted to an ammeter by connecting a small shunt resistance S in parallel (see figure). Kirchhoff's loop law gives,
\({{i}_{g}}G-(i-{{i}_{g}})S=0,\Rightarrow i={{i}_{g}}(G+S)/S\)
The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter. Substitute the values to get
\(i={{i}_{g}}(G+S)/S=(100\times {{10}^{-6}})((100+0.1)/0.1))=100.1mA.\)
1. 2 moving coil galvanometers M1 and M2(having the same spring constants), have
the following specifications:
R1 = 10Ω, N1 = 30, A1 = 3.6×10-3m2, B1 = 0.25T
R2 = 14Ω, N2 = 42, A2 = 1.8×10-3m2, B2 = 0.5T
Find the ratios of current sensitivities between the 2 galvanometers.
Solution:
current sensitivity is given by:
Φ/i = NAB/k
(Φ/i)1 = (30×3.6×10-3×0.25)/k
(Φ/i)2 = (42×1.8×10-3×0.5)/k
∴(Φ/i)1/(Φ/i)2 = (30×3.6×10-3×0.25)/(42×1.8×10-3×0.5) = 5/7
2. 2 moving coil galvanometers M1 and M2(having the same spring constants),
have the following specifications:
R1 = 10Ω, N1 = 30, A1 = 3.6×10-3m2, B1 = 0.25T
R2 = 14Ω, N2 = 42, A2 = 1.8×10-3m2, B2 = 0.5T
Find the ratios of Voltage sensitivities between the 2 galvanometers.
Solution:
Φ/V = NAB/(KR)
(Φ/V)1 = (30×3.6×10-3×0.25)/(k×10)
(Φ/V)2 = (42×1.8×10-3×0.5)/(k×14)
∴ (Φ/V)1/(Φ/V)2 = (30×3.6×10-3×0.25×14)/(42×1.8×10-3×0.5×10) = 1
A magnetic field line or magnetic flux line shows the direction of a magnet's force and the strength of a magnet.
One can make magnetic field lines show as if they were physical phenomena. For example, iron filings placed in a magnetic field line up to form lines that correspond to 'field lines'.
If there are a lot of lines through a magnet and not a great space between them, the magnet is strong. If the lines between a magnet are far apart and there aren't many lines, the magnet is weak. A way of determining the strength of a magnet is to do an experiment with iron filings. The iron filings will get attracted to the magnet and move into the shape of the flux lines. You then look at the shape of the iron filings and see the gap between the flux lines. This gives you an idea of the strength of the magnet.
The use of iron filings to display a field alters the magnetic field so that it is much larger along the "lines" of iron. This caused by the large permeability of iron relative to air. Magnetic fields' "lines" are also visually displayed in polar auroras, when particles cause visible streaks of light that line up with the local direction of Earth's magnetic field.
Magnetic field lines are like the contour lines (constant altitude) on a topographic map in that they represent something continuous, and a different mapping scale would show more or fewer lines. There is an advantage to using magnetic field lines as a representation. Many laws of magnetism (and electromagnetism) can be stated completely and concisely using simple concepts such as the 'number' of field lines through a surface. These concepts can be quickly 'translated' to their mathematical form.
Magnetic Dipole:
generally a tiny magnet of microscopic to subatomic dimensions, equivalent to a flow of electric charge around a loop. Electrons circulating around atomic nuclei, electrons spinning on their axes, and rotating positively charged atomic nuclei all are magnetic dipoles. The sum of these effects may cancel so that a given type of atom may not be a magnetic dipole. If they do not fully cancel, the atom is a permanent magnetic dipole, as are iron atoms. Many millions of iron atoms spontaneously locked into the same alignment to form a ferromagnetic domain also constitute a magnetic dipole. Magnetic compass needles and bar magnets are examples of macroscopic magnetic dipoles.
Magnetic Dipole Moment :
If current is flowing in a coil it behaves like a magnet. If the current if flowing in anticlockwise direction it behaves like a north pole and if the current is flowing in clockwise direction it behaves like a south pole.
Magnetic dipole moment gives the effectiveness of a magnet.
Magnetic dipole moment depends upon the pole strength.
Magnetic dipole moment creates magnetic field.
Magnetic field at a distance ‘r’ from the center of bar magnet is -
\(B=k\frac{2M}{{{r}^{3}}}\)
Where,
M = magnetic dipole moment of bar magnet.
Magnetic field at a distance ‘r’ from the center of current carrying circular coil is -
\(B=k\frac{2M}{{{r}^{3}}}\)
Where,
M = magnetic dipole moment of current carrying circular coil.
The product of current (I) flowing in the loop and the area (A) of the loop is called magnetic dipole moment (μ),
\(\mu =N\vec{I}\vec{A}\)
Where,
N= number of turns
Where direction of area vector will be same as that of the direction of magnetic field at the center of loop.
Solution:
Magnetic dipole moment of coil = NIA
Solution:
An end-on position
1. A magnetic dipole is under the influence of two magnetic fields having an angle of 1200 between them. One of the fields has a magnitude . If the dipole comes to stable equilibrium at an angle of 300 with this field, then magnitude of the other field is
Solution:
\({{B}_{1}}\sin {{30}^{0}}={{B}_{2}}\sin {{90}^{0}}\)
= 0.6 x10-2 T
2. A short bar magnet is placed with its south pole facing geographic south and the distance between the null points is found to be 16 cm. When the magnet is turned pole to pole at the same place then the distance between the null points will be
Solution:
\({{B}_{H}}={{B}_{{{e}_{1}}}}=\frac{{{\mu }_{0}}2M}{4\pi d_{1}^{3}}\)
\({{B}_{H}}={{B}_{{{e}_{2}}}}=\frac{{{\mu }_{0}}M}{4\pi d_{2}^{3}}\)
will be 16 21/3 times on the axial line
A uniformly charged disc whose total charge has magnitude q and whose radius is r rotates with constant angular velocity of magnitude w. What is the magnetic dipole moment?
Solution:
The surface charge density is q/pr2. Hence the charge within a ring of radius R and width dR is
\(dq=\frac{q}{\pi {{r}^{2}}}(2\pi RdR)=\frac{2q}{{{r}^{2}}}(RdR)\)
The current carried by this ring is its charge divided by the rotation period,
\(di=\frac{dq}{2\pi /\omega }=\frac{q\omega }{\pi {{r}^{2}}}[R.dR]\)
The magnetic moment contributed by this ring has the magnitude dM = a |di|, where a is the area of the ring.
dM = pR2 |di| = q. w/pr2 (RdR)
\(M=\int_{{}}^{{}}{dM=\int_{\,r=0}^{\,r}{q.\frac{\omega }{{{r}^{2}}}({{R}^{3}}dR)=q\omega \frac{{{r}^{2}}}{4}}}\)
A magnetic field line or magnetic flux line shows the direction of a magnet's force and the strength of a magnet.
One can make magnetic field lines show as if they were physical phenomena. For example, iron filings placed in a magnetic field line up to form lines that correspond to 'field lines'.
If there are a lot of lines through a magnet and not a great space between them, the magnet is strong. If the lines between a magnet are far apart and there aren't many lines, the magnet is weak. A way of determining the strength of a magnet is to do an experiment with iron filings. The iron filings will get attracted to the magnet and move into the shape of the flux lines. You then look at the shape of the iron filings and see the gap between the flux lines. This gives you an idea of the strength of the magnet.
The use of iron filings to display a field alters the magnetic field so that it is much larger along the "lines" of iron. This caused by the large permeability of iron relative to air. Magnetic fields' "lines" are also visually displayed in polar auroras, when particles cause visible streaks of light that line up with the local direction of Earth's magnetic field.
Magnetic field lines are like the contour lines (constant altitude) on a topographic map in that they represent something continuous, and a different mapping scale would show more or fewer lines. There is an advantage to using magnetic field lines as a representation. Many laws of magnetism (and electromagnetism) can be stated completely and concisely using simple concepts such as the 'number' of field lines through a surface. These concepts can be quickly 'translated' to their mathematical form.
Magnetic Dipole:
generally a tiny magnet of microscopic to subatomic dimensions, equivalent to a flow of electric charge around a loop. Electrons circulating around atomic nuclei, electrons spinning on their axes, and rotating positively charged atomic nuclei all are magnetic dipoles. The sum of these effects may cancel so that a given type of atom may not be a magnetic dipole. If they do not fully cancel, the atom is a permanent magnetic dipole, as are iron atoms. Many millions of iron atoms spontaneously locked into the same alignment to form a ferromagnetic domain also constitute a magnetic dipole. Magnetic compass needles and bar magnets are examples of macroscopic magnetic dipoles.
Magnetic Dipole Moment :
If current is flowing in a coil it behaves like a magnet. If the current if flowing in anticlockwise direction it behaves like a north pole and if the current is flowing in clockwise direction it behaves like a south pole.
Magnetic dipole moment gives the effectiveness of a magnet.
Magnetic dipole moment depends upon the pole strength.
Magnetic dipole moment creates magnetic field.
Magnetic field at a distance ‘r’ from the center of bar magnet is -
\(B=k\frac{2M}{{{r}^{3}}}\)
Where,
M = magnetic dipole moment of bar magnet.
Magnetic field at a distance ‘r’ from the center of current carrying circular coil is -
\(B=k\frac{2M}{{{r}^{3}}}\)
Where,
M = magnetic dipole moment of current carrying circular coil.
The product of current (I) flowing in the loop and the area (A) of the loop is called magnetic dipole moment (μ),
\(\mu =N\vec{I}\vec{A}\)
Where,
N= number of turns
Where direction of area vector will be same as that of the direction of magnetic field at the center of loop.
Solution:
Magnetic dipole moment of coil = NIA
Solution:
An end-on position
1. A magnetic dipole is under the influence of two magnetic fields having an angle of 1200 between them. One of the fields has a magnitude . If the dipole comes to stable equilibrium at an angle of 300 with this field, then magnitude of the other field is
Solution:
\({{B}_{1}}\sin {{30}^{0}}={{B}_{2}}\sin {{90}^{0}}\)
= 0.6 x10-2 T
2. A short bar magnet is placed with its south pole facing geographic south and the distance between the null points is found to be 16 cm. When the magnet is turned pole to pole at the same place then the distance between the null points will be
Solution:
\({{B}_{H}}={{B}_{{{e}_{1}}}}=\frac{{{\mu }_{0}}2M}{4\pi d_{1}^{3}}\)
\({{B}_{H}}={{B}_{{{e}_{2}}}}=\frac{{{\mu }_{0}}M}{4\pi d_{2}^{3}}\)
will be 16 21/3 times on the axial line
A uniformly charged disc whose total charge has magnitude q and whose radius is r rotates with constant angular velocity of magnitude w. What is the magnetic dipole moment?
Solution:
The surface charge density is q/pr2. Hence the charge within a ring of radius R and width dR is
\(dq=\frac{q}{\pi {{r}^{2}}}(2\pi RdR)=\frac{2q}{{{r}^{2}}}(RdR)\)
The current carried by this ring is its charge divided by the rotation period,
\(di=\frac{dq}{2\pi /\omega }=\frac{q\omega }{\pi {{r}^{2}}}[R.dR]\)
The magnetic moment contributed by this ring has the magnitude dM = a |di|, where a is the area of the ring.
dM = pR2 |di| = q. w/pr2 (RdR)
\(M=\int_{{}}^{{}}{dM=\int_{\,r=0}^{\,r}{q.\frac{\omega }{{{r}^{2}}}({{R}^{3}}dR)=q\omega \frac{{{r}^{2}}}{4}}}\)
Coulomb's law, or Coulomb's inverse-square law, is a law of physics for quantifying the amount of force with which stationary electrically charged particles repel or attract each other. In its scalar form, the law is:
\(F={{k}_{e}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}},\)
where ke is Coulomb's constant, q1 and q2 are the signed magnitudes of the charges, and the scalar r is the distance between the charges. The force of the interaction between the charges is attractive if the charges have opposite signs (i.e., Fis negative) and repulsive if like-signed (i.e., F is positive).
The law was essential to the development of the theory of electromagnetism. Being an inverse-square law, it is analogous to Isaac Newton's inverse-square law of universal gravitation. Coulomb's law can be used to derive Gauss's law, and vice versa. The law has been tested extensively, and all observations have upheld the law's principle.
Coulomb's law:
The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
The force is along the straight line joining them. If the two charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.
Coulomb's law can also be stated as a simple mathematical expression. The scalar and vector forms of the mathematical equation are
\(\left| F \right|={{k}_{e}}\frac{\left| {{q}_{1}}{{q}_{2}} \right|}{{{r}^{2}}}\) and \({{F}_{1}}={{k}_{e}}\frac{{{q}_{1}}{{q}_{2}}}{{{\left| {{r}_{21}} \right|}^{2}}}{{\hat{r}}_{21,}}\) respectively
where ke is Coulomb's constant, q1 and q2 are the signed magnitudes of the charges, the scalar r is the distance between the charges, the vector r21 = r1 − r2 is the vectorial distance between the charges, and r̂21 = r21/|r21| (a unit vector pointing from q2 to q1). The vector form of the equation calculates the force F1 applied on q1 by q2. If r12 is used instead, then the effect on q2 can be found. It can be also calculated using Newton's third law: F2 = −F1.
Coulomb's constant:
Coulomb's constant is a proportionality factor that appears in Coulomb's law as well as in other electric-related formulas. Denoted ke, it is also called the electric force constant or electrostatic constant, hence the subscript e.
The exact value of Coulomb's constant is :
\(\begin{align} & {{K}_{e}}=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{{{C}_{0}}^{2}{{\mu }_{0}}}{4\pi }={{C}_{0}}^{2}\times {{10}^{-7}}H.{{m}^{-1}} \\ & =8.9875517873681764\times {{10}^{9}}N.{{m}^{2}}.{{C}^{-2}} \\ \end{align}\)
Simple Experiment to Verify Coulomb's Law:
Consider two small spheres of mass m and same-sign charge q, hanging from two ropes of negligible mass of length l. The forces acting on each sphere are three: the weight mg, rope tension T and the electric force F.
In the equilibrium state:
\(T=\sin {{\theta }_{1}}={{F}_{1}}\)
and:
\(T\cos \theta =mg\)
Dividing (1) by (2),
\(\frac{\sin {{\theta }_{1}}}{\cos {{\theta }_{1}}}=\frac{{{F}_{1}}}{mg}\Rightarrow {{F}_{1}}=mg\tan {{\theta }_{1}}\)
Let L1 be the distance between the charged spheres; the repulsion force between them F1, assuming Coulomb's law is correct, is equal to
\({{F}_{1}}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}L_{1}^{2}}\) (Coulomb's law)
So,
\(\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}L_{1}^{2}}=mg\tan {{\theta }_{1}}\)
If we now discharge one of the spheres, and we put it in contact with the charged sphere, each one of them acquires a charge \(\frac{q}{2}\).In the equilibrium state, the distance between the charges will be L2 < L1 and the repulsion force between them will be:
\({{F}_{2}}=\frac{{{\left( \frac{q}{2} \right)}^{2}}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}}=\frac{\frac{{{q}^{2}}}{4}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}}\)
We know that \({{F}_{2}}=mg\tan {{\theta }_{2}}\). And:
\(\frac{\frac{{{q}^{2}}}{4}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}}=mg\tan {{\theta }_{2}}\)
Dividing (4) by (5), we get
\(\frac{\left( \frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}L_{1}^{2}} \right)}{\left( \frac{\frac{{{q}^{2}}}{4}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}} \right)}=\frac{mg\tan {{\theta }_{1}}}{mg\tan {{\theta }_{2}}}\Rightarrow 4{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\)
Measuring the angles \({{\theta }_{1}}\) and \({{\theta }_{2}}\) and the distance between the charges L1 and L2 is sufficient to verify that the equality is true taking into account the experimental error. In practice, angles can be difficult to measure, so if the length of the ropes is sufficiently great, the angles will be small enough to make the following approximation:
\(\tan \theta \approx \sin \theta =\frac{\frac{L}{2}}{l}=\frac{L}{2l}\Rightarrow \frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\approx \frac{\frac{{{L}_{1}}}{2l}}{\frac{{{L}_{2}}}{2l}}\)
Using this approximation, the relationship (6) becomes the much simpler expression:
\(\frac{\frac{{{L}_{1}}}{2l}}{\frac{{{L}_{2}}}{2l}}\approx 4{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}\Rightarrow \frac{{{L}_{1}}}{{{L}_{2}}}\approx 4{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}\Rightarrow \frac{{{L}_{1}}}{{{L}_{2}}}\approx \sqrt[3]{4}\)
In this way, the verification is limited to measuring the distance between the charges and check that the division approximates the theoretical value.
1. Two charged objects have a repulsive force of 0.080 N. If the distance separating the objects is tripled, then what is the new force?
Solution:
The electrostatic force is inversely related to the square of the separation distance. So if d is three times larger, then F is nine times smaller - that is, one-ninth the original value. One-ninth of 0.080 N is 0.00889 N.
2. Two charged objects have a repulsive force of 0.080 N. If the distance separating the objects is halved, then what is the new force?
Solution:
The electrostatic force is inversely related to the square of the separation distance. So if d is two times smaller, then F is four times larger. Four times 0.080 N is 0.320 N
1. Determine the electrical force of attraction between two balloons with separate charges of +3.5 x 10-8C and -2.9 x 10-8 C when separated a distance of 0.65 m.
Solution:
Step 1: Identify known values in variable form.
Q1 = +3.5 x 10-8 C and Q2 = -2.9 x 10-8 C
d = 0.65 m
Step 2: Identify the requested value
F = ???
Step 3: Substitute and solve.
2. Joann has rubbed a balloon with wool to give it a charge of -1.0 x 10-6 C. She then acquires a plastic golf tube with a charge of +4.0 x 10-6 C localized at a given position. She holds the location of charge on the plastic golf tube a distance of 50.0 cm above the balloon. Determine the electrical force of attraction between the golf tube and the balloon.
Solution:
Step 1: Identify known values in variable form.
Q1 = -1.0 x 10^-6 C and Q2 = +4.0 x 10-6 C
d = 50.0 cm = 0.50 m.
Step 2: Identify requested information
F = ???
Step 3: Substitute and solve.
1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C−2 = 9 x 109 Nm2C−2.
Charge A (qA) = +8 μC = +8 x 10-6 C
Charge B (qB) = -5 μC = -5 x 10-6 C
k = 9 x 109 Nm2C−2
The distance between charge A and B (rAB) = 10 cm = 0.1 m
Find- The magnitude of the electric force
Solution:
Formula of Coulomb’s law :
\(F=k\frac{{{q}_{A}}{{q}_{B}}}{{{r}_{AB}}}\)
The magnitude of the electric force :
\(\begin{align} & F=\frac{(8\times {{10}^{-6}})(5\times {{10}^{-6}})}{{{(0.1)}^{2}}} \\ & F=\frac{40\times {{10}^{-12}}}{{{(0.1)}^{2}}}=\frac{40\times {{10}^{-12}}}{0.01}=4000\times {{10}^{-12}} \\ & F=4\times {{10}^{-9}}N \\ \end{align}\)
1. Four charges are located on the corners of a square. the sides of the square have a length of 0.05m the upper left corner has a charge of -q. the upper right corner has a charge of -q. the lower left corner has a charge of 2q. the lower right corner has a charge of -2q. if the magnitude of q is \(1.0\times {{10}^{-7}}C\), find the magnitude of the force exerted on the charge at the lower left corner of this system.
Solution:
No. of charges: lower left = 1, upper left = 2, upper right = 3, lower right = 4
Distance = 0.05 m
\(\begin{align} & {{r}_{1,2}}={{r}_{1,3}}=(\sqrt{2})a,{{r}_{1,4}}=a \\ & {{F}_{1,2}}=k(2q)(q)/{{a}^{2}}=0.072N(repulsive) \\ & {{F}_{1,3}}=k(2q)(q)/{{(\sqrt{2}a)}^{2}}=0.036N(attractive,\frac{{}}{{}}at\frac{{}}{{}}a\frac{{}}{{}}{{40}^{o}}angle \\ & {{F}_{1,4}}=k(2q)(2q)/{{a}^{2}}=0.144N(attractive) \\ & \sum{{{F}_{x}}}={{F}_{1,3}}\cos (45)+{{F}_{1,4}}=0.036\cos 45+0.144=0.17N \\ & \sum{{{F}_{y}}}={{F}_{1,3}}\sin (45)-{{F}_{1,2}}=-0.05N \\ & F=0.18N \\ \end{align}\)
Coulomb's law, or Coulomb's inverse-square law, is a law of physics for quantifying the amount of force with which stationary electrically charged particles repel or attract each other. In its scalar form, the law is:
\(F={{k}_{e}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}},\)
where ke is Coulomb's constant, q1 and q2 are the signed magnitudes of the charges, and the scalar r is the distance between the charges. The force of the interaction between the charges is attractive if the charges have opposite signs (i.e., Fis negative) and repulsive if like-signed (i.e., F is positive).
The law was essential to the development of the theory of electromagnetism. Being an inverse-square law, it is analogous to Isaac Newton's inverse-square law of universal gravitation. Coulomb's law can be used to derive Gauss's law, and vice versa. The law has been tested extensively, and all observations have upheld the law's principle.
Coulomb's law:
The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
The force is along the straight line joining them. If the two charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.
Coulomb's law can also be stated as a simple mathematical expression. The scalar and vector forms of the mathematical equation are
\(\left| F \right|={{k}_{e}}\frac{\left| {{q}_{1}}{{q}_{2}} \right|}{{{r}^{2}}}\) and \({{F}_{1}}={{k}_{e}}\frac{{{q}_{1}}{{q}_{2}}}{{{\left| {{r}_{21}} \right|}^{2}}}{{\hat{r}}_{21,}}\) respectively
where ke is Coulomb's constant, q1 and q2 are the signed magnitudes of the charges, the scalar r is the distance between the charges, the vector r21 = r1 − r2 is the vectorial distance between the charges, and r̂21 = r21/|r21| (a unit vector pointing from q2 to q1). The vector form of the equation calculates the force F1 applied on q1 by q2. If r12 is used instead, then the effect on q2 can be found. It can be also calculated using Newton's third law: F2 = −F1.
Coulomb's constant:
Coulomb's constant is a proportionality factor that appears in Coulomb's law as well as in other electric-related formulas. Denoted ke, it is also called the electric force constant or electrostatic constant, hence the subscript e.
The exact value of Coulomb's constant is :
\(\begin{align} & {{K}_{e}}=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{{{C}_{0}}^{2}{{\mu }_{0}}}{4\pi }={{C}_{0}}^{2}\times {{10}^{-7}}H.{{m}^{-1}} \\ & =8.9875517873681764\times {{10}^{9}}N.{{m}^{2}}.{{C}^{-2}} \\ \end{align}\)
Simple Experiment to Verify Coulomb's Law:
Consider two small spheres of mass m and same-sign charge q, hanging from two ropes of negligible mass of length l. The forces acting on each sphere are three: the weight mg, rope tension T and the electric force F.
In the equilibrium state:
\(T=\sin {{\theta }_{1}}={{F}_{1}}\)
and:
\(T\cos \theta =mg\)
Dividing (1) by (2),
\(\frac{\sin {{\theta }_{1}}}{\cos {{\theta }_{1}}}=\frac{{{F}_{1}}}{mg}\Rightarrow {{F}_{1}}=mg\tan {{\theta }_{1}}\)
Let L1 be the distance between the charged spheres; the repulsion force between them F1, assuming Coulomb's law is correct, is equal to
\({{F}_{1}}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}L_{1}^{2}}\) (Coulomb's law)
So,
\(\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}L_{1}^{2}}=mg\tan {{\theta }_{1}}\)
If we now discharge one of the spheres, and we put it in contact with the charged sphere, each one of them acquires a charge \(\frac{q}{2}\).In the equilibrium state, the distance between the charges will be L2 < L1 and the repulsion force between them will be:
\({{F}_{2}}=\frac{{{\left( \frac{q}{2} \right)}^{2}}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}}=\frac{\frac{{{q}^{2}}}{4}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}}\)
We know that \({{F}_{2}}=mg\tan {{\theta }_{2}}\). And:
\(\frac{\frac{{{q}^{2}}}{4}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}}=mg\tan {{\theta }_{2}}\)
Dividing (4) by (5), we get
\(\frac{\left( \frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}L_{1}^{2}} \right)}{\left( \frac{\frac{{{q}^{2}}}{4}}{4\pi {{\varepsilon }_{0}}L_{2}^{2}} \right)}=\frac{mg\tan {{\theta }_{1}}}{mg\tan {{\theta }_{2}}}\Rightarrow 4{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\)
Measuring the angles \({{\theta }_{1}}\) and \({{\theta }_{2}}\) and the distance between the charges L1 and L2 is sufficient to verify that the equality is true taking into account the experimental error. In practice, angles can be difficult to measure, so if the length of the ropes is sufficiently great, the angles will be small enough to make the following approximation:
\(\tan \theta \approx \sin \theta =\frac{\frac{L}{2}}{l}=\frac{L}{2l}\Rightarrow \frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\approx \frac{\frac{{{L}_{1}}}{2l}}{\frac{{{L}_{2}}}{2l}}\)
Using this approximation, the relationship (6) becomes the much simpler expression:
\(\frac{\frac{{{L}_{1}}}{2l}}{\frac{{{L}_{2}}}{2l}}\approx 4{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}\Rightarrow \frac{{{L}_{1}}}{{{L}_{2}}}\approx 4{{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}\Rightarrow \frac{{{L}_{1}}}{{{L}_{2}}}\approx \sqrt[3]{4}\)
In this way, the verification is limited to measuring the distance between the charges and check that the division approximates the theoretical value.
1. Two charged objects have a repulsive force of 0.080 N. If the distance separating the objects is tripled, then what is the new force?
Solution:
The electrostatic force is inversely related to the square of the separation distance. So if d is three times larger, then F is nine times smaller - that is, one-ninth the original value. One-ninth of 0.080 N is 0.00889 N.
2. Two charged objects have a repulsive force of 0.080 N. If the distance separating the objects is halved, then what is the new force?
Solution:
The electrostatic force is inversely related to the square of the separation distance. So if d is two times smaller, then F is four times larger. Four times 0.080 N is 0.320 N
1. Determine the electrical force of attraction between two balloons with separate charges of +3.5 x 10-8C and -2.9 x 10-8 C when separated a distance of 0.65 m.
Solution:
Step 1: Identify known values in variable form.
Q1 = +3.5 x 10-8 C and Q2 = -2.9 x 10-8 C
d = 0.65 m
Step 2: Identify the requested value
F = ???
Step 3: Substitute and solve.
2. Joann has rubbed a balloon with wool to give it a charge of -1.0 x 10-6 C. She then acquires a plastic golf tube with a charge of +4.0 x 10-6 C localized at a given position. She holds the location of charge on the plastic golf tube a distance of 50.0 cm above the balloon. Determine the electrical force of attraction between the golf tube and the balloon.
Solution:
Step 1: Identify known values in variable form.
Q1 = -1.0 x 10^-6 C and Q2 = +4.0 x 10-6 C
d = 50.0 cm = 0.50 m.
Step 2: Identify requested information
F = ???
Step 3: Substitute and solve.
1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C−2 = 9 x 109 Nm2C−2.
Charge A (qA) = +8 μC = +8 x 10-6 C
Charge B (qB) = -5 μC = -5 x 10-6 C
k = 9 x 109 Nm2C−2
The distance between charge A and B (rAB) = 10 cm = 0.1 m
Find- The magnitude of the electric force
Solution:
Formula of Coulomb’s law :
\(F=k\frac{{{q}_{A}}{{q}_{B}}}{{{r}_{AB}}}\)
The magnitude of the electric force :
\(\begin{align} & F=\frac{(8\times {{10}^{-6}})(5\times {{10}^{-6}})}{{{(0.1)}^{2}}} \\ & F=\frac{40\times {{10}^{-12}}}{{{(0.1)}^{2}}}=\frac{40\times {{10}^{-12}}}{0.01}=4000\times {{10}^{-12}} \\ & F=4\times {{10}^{-9}}N \\ \end{align}\)
1. Four charges are located on the corners of a square. the sides of the square have a length of 0.05m the upper left corner has a charge of -q. the upper right corner has a charge of -q. the lower left corner has a charge of 2q. the lower right corner has a charge of -2q. if the magnitude of q is \(1.0\times {{10}^{-7}}C\), find the magnitude of the force exerted on the charge at the lower left corner of this system.
Solution:
No. of charges: lower left = 1, upper left = 2, upper right = 3, lower right = 4
Distance = 0.05 m
\(\begin{align} & {{r}_{1,2}}={{r}_{1,3}}=(\sqrt{2})a,{{r}_{1,4}}=a \\ & {{F}_{1,2}}=k(2q)(q)/{{a}^{2}}=0.072N(repulsive) \\ & {{F}_{1,3}}=k(2q)(q)/{{(\sqrt{2}a)}^{2}}=0.036N(attractive,\frac{{}}{{}}at\frac{{}}{{}}a\frac{{}}{{}}{{40}^{o}}angle \\ & {{F}_{1,4}}=k(2q)(2q)/{{a}^{2}}=0.144N(attractive) \\ & \sum{{{F}_{x}}}={{F}_{1,3}}\cos (45)+{{F}_{1,4}}=0.036\cos 45+0.144=0.17N \\ & \sum{{{F}_{y}}}={{F}_{1,3}}\sin (45)-{{F}_{1,2}}=-0.05N \\ & F=0.18N \\ \end{align}\)
(A) If a magnet of length l and magnetic moment M gets bent in the form of a semicircular, then its new magnetic moment will be M’ = 2M/π
(B) The magnetic moment of a given electron because of its orbital motion is 1μB. But due to its spin motion, it will be μB /2.
\(\begin{align} & i.e.{{M}_{orbital}}=l\left( \frac{eh}{4\pi {{m}_{1}}} \right)=\left( \frac{\bar{e}h}{2{{m}_{1}}} \right) \\ & and{{M}_{spin}}=s\left( \frac{eh}{4\pi {{m}_{s}}} \right)=S\left( \frac{\bar{e}h}{2{{m}_{s}}} \right)=S{{\mu }_{B}}=\frac{{{\mu }_{B}}}{2} \\ \end{align}\)
Here μB = Bohr magneton
(i) The value of Bohr magneton μB = eh / 4πm
(ii) μB = 0.93 × 10–23 Amp-m2
(c) Other formulae for magnetic moments:
(i) M = ni r2
(ii) M = eVr/2 = er2ω/2 = er22πf/2 = er2π/T
(iii) e/mJ
(iv) M = nωB
(d) How can we calculate resultant magnetic moment?
(i) When two bar magnets are lying mutually perpendicular to each other, then
M = √M12 + M22 = √2mpl
(ii) When two coils with a radius of r and carrying current I each, are lying concentrically with their planes at right angles to each other, then
M = √M12 + M22 = √2iπr2 if M1 = M2
The Earth's magnetic field is similar to that of a bar magnet tilted 11 degrees from the spin axis of the Earth. The problem with that picture is that the Curie temperature of iron is about 770 C . The Earth's core is hotter than that and therefore not magnetic. So how did the Earth get its magnetic field?
Magnetic fields surround electric curents, so we surmise that circulating electric currents in the Earth's molten metallic core are the origin of the magnetic field. A curent loop gives a field similar to that of the earth. The magnetic field magnitude measured at the surface of the Earth is about half a Gauss and dips toward the Earth in the northern hemisphere. The magnitude varies over the surface of the Earth in the range 0.3 to 0.6 Gauss.
The Earth's magnetic field is attributed to a dynamo effect of circulating electric current, but it is not constant in direction. Rock specimens of different age in similar locations have different directions of permanent magnetization. Evidence for 171 magnetic field reversals during the past 71 million years has been reported.
Although the details of the dynamo effect are not known in detail, the rotation of the Earth plays a part in generating the currents which are presumed to be the source of the magnetic field. Mariner 2 found that Venus does not have such a magnetic field although its core iron content must be similar to that of the Earth. Venus's rotation period of 243 Earth days is just too slow to produce the dynamo effect.
Interaction of the terrestrial magnetic field with particles from the solar wind sets up the conditions for the aurora phenomena near the poles.
The north pole of a compass needle is a magnetic north pole. It is attracted to the geographic North Pole, which is a magnetic south pole (opposite magnetic poles attract).
When a material is placed within a magnetic field, the magnetic forces of the material's electrons will be affected. This effect is known as Faraday's Law of Magnetic Induction. However, materials can react quite differently to the presence of an external magnetic field. This reaction is dependent on a number of factors, such as the atomic and molecular structure of the material, and the net magnetic field associated with the atoms. The magnetic moments associated with atoms have three origins. These are the electron motion, the change in motion caused by an external magnetic field, and the spin of the electrons.
In most atoms, electrons occur in pairs. Electrons in a pair spin in opposite directions. So, when electrons are paired together, their opposite spins cause their magnetic fields to cancel each other. Therefore, no net magnetic field exists. Alternately, materials with some unpaired electrons will have a net magnetic field and will react more to an external field. Most materials can be classified as diamagnetic, paramagnetic or ferromagnetic.
Diamagnetic:
Diamagnetic materials have a weak, negative susceptibility to magnetic fields. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. Diamagnetic properties arise from the realignment of the electron paths under the influence of an external magnetic field. Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.
If we plot M vs H, we see:
Note that when the field is zero the magnetization is zero. The other characteristic behavior of diamagnetic materials is that the susceptibility is temperature independent. Some well known diamagnetic substances, in units of 10-8 m3/kg, include:
quartz (SiO2) -0.62
Calcite (CaCO3) -0.48
water -0.90
Paramagnetic:
Paramagnetic materials have a small, positive susceptibility to magnetic fields. These materials are slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.
In addition, the efficiency of the field in aligning the moments is opposed by the randomizing effects of temperature. This results in a temperature dependent susceptibility, known as the Curie Law.
At normal temperatures and in moderate fields, the paramagnetic susceptibility is small (but larger than the diamagnetic contribution). Unless the temperature is very low (<<100 K) or the field is very high paramagnetic susceptibility is independent of the applied field. Under these conditions, paramagnetic susceptibility is proportional to the total iron content. Many iron bearing minerals are paramagnetic at room temperature. Some examples, in units of 10-8 m3/kg, include:
Montmorillonite (clay) 13
Nontronite (Fe-rich clay) 65
Biotite (silicate) 79
Siderite(carbonate) 100
Pyrite (sulfide) 30
The paramagnetism of the matrix minerals in natural samples can be significant if the concentration of magnetite is very small. In this case, a paramagnetic correction may be needed.
Ferromagnetic:
Ferromagnetic materials have a large, positive susceptibility to an external magnetic field. They exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. Ferromagnetic materials have some unpaired electrons so their atoms have a net magnetic moment. They get their strong magnetic properties due to the presence of magnetic domains. In these domains, large numbers of atom's moments (1012 to 1015) are aligned parallel so that the magnetic force within the domain is strong. When a ferromagnetic material is in the unmagnitized state, the domains are nearly randomly organized and the net magnetic field for the part as a whole is zero. When a magnetizing force is applied, the domains become aligned to produce a strong magnetic field within the part. Iron, nickel, and cobalt are examples of ferromagnetic materials. Components with these materials are commonly inspected using the magnetic particle method.
Ferromagnetic materials exhibit parallel alignment of moments resulting in large net magnetization even in the absence of a magnetic field. The elements Fe, Ni, and Co and many of their alloys are typical ferromagnetic materials.
Two distinct characteristics of ferromagnetic materials are their
(1) spontaneous magnetization and the existence of
(2) magnetic ordering temperature
Solution:
Repelled due to induction of similar poles.
Solution:
cm is positive at all temperature
Solution:
It will Fall, because diamagnetic substance, moves from stronger magnetic field to weaker field.
1. A square loop OABCO of side l carries a current i. Now, find the magnetic moment of the loop.
Solution:
Magnetic moment of the loop can be written as,
M = i (BC x CO), where the letters in bold denote the vectors
BC = -lk
1. Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i = 2.0 A.
Solution:
By assuming two equal and opposite currents in BE, two current carrying loop (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence
Mnet = √M2 + M2 = √2 M
Where M = ia = (2.0)(0.1)(0.1) = 0.02 A-m2
Mnet = (√2) (0.02) A – m2 = 0.028 A-m2
2. Two identical bar magnets each of length L and pole–strength m are placed at right angles to each other with the north pole of one touching the south pole of the other. Evaluate the magnetic moment of the system.
Solution:
As magnetic moment is a vector, so we know
MR = (M12 + M22 + 2M1M2 cos θ)1/2 with tanΦ M2 sinθ / M1 + M2 cosθ
and as here M1= M2 = mL and θ = 90°, so
MR = (M2 + M2 + 2MM cos 90)1/2 = (√2) mL
And, tanΦ Msin90 / M + Mcos 90 = 1, i.e., Φ = tan–1(1) = 45°
(A) If a magnet of length l and magnetic moment M gets bent in the form of a semicircular, then its new magnetic moment will be M’ = 2M/π
(B) The magnetic moment of a given electron because of its orbital motion is 1μB. But due to its spin motion, it will be μB /2.
\(\begin{align} & i.e.{{M}_{orbital}}=l\left( \frac{eh}{4\pi {{m}_{1}}} \right)=\left( \frac{\bar{e}h}{2{{m}_{1}}} \right) \\ & and{{M}_{spin}}=s\left( \frac{eh}{4\pi {{m}_{s}}} \right)=S\left( \frac{\bar{e}h}{2{{m}_{s}}} \right)=S{{\mu }_{B}}=\frac{{{\mu }_{B}}}{2} \\ \end{align}\)
Here μB = Bohr magneton
(i) The value of Bohr magneton μB = eh / 4πm
(ii) μB = 0.93 × 10–23 Amp-m2
(c) Other formulae for magnetic moments:
(i) M = ni r2
(ii) M = eVr/2 = er2ω/2 = er22πf/2 = er2π/T
(iii) e/mJ
(iv) M = nωB
(d) How can we calculate resultant magnetic moment?
(i) When two bar magnets are lying mutually perpendicular to each other, then
M = √M12 + M22 = √2mpl
(ii) When two coils with a radius of r and carrying current I each, are lying concentrically with their planes at right angles to each other, then
M = √M12 + M22 = √2iπr2 if M1 = M2
The Earth's magnetic field is similar to that of a bar magnet tilted 11 degrees from the spin axis of the Earth. The problem with that picture is that the Curie temperature of iron is about 770 C . The Earth's core is hotter than that and therefore not magnetic. So how did the Earth get its magnetic field?
Magnetic fields surround electric curents, so we surmise that circulating electric currents in the Earth's molten metallic core are the origin of the magnetic field. A curent loop gives a field similar to that of the earth. The magnetic field magnitude measured at the surface of the Earth is about half a Gauss and dips toward the Earth in the northern hemisphere. The magnitude varies over the surface of the Earth in the range 0.3 to 0.6 Gauss.
The Earth's magnetic field is attributed to a dynamo effect of circulating electric current, but it is not constant in direction. Rock specimens of different age in similar locations have different directions of permanent magnetization. Evidence for 171 magnetic field reversals during the past 71 million years has been reported.
Although the details of the dynamo effect are not known in detail, the rotation of the Earth plays a part in generating the currents which are presumed to be the source of the magnetic field. Mariner 2 found that Venus does not have such a magnetic field although its core iron content must be similar to that of the Earth. Venus's rotation period of 243 Earth days is just too slow to produce the dynamo effect.
Interaction of the terrestrial magnetic field with particles from the solar wind sets up the conditions for the aurora phenomena near the poles.
The north pole of a compass needle is a magnetic north pole. It is attracted to the geographic North Pole, which is a magnetic south pole (opposite magnetic poles attract).
When a material is placed within a magnetic field, the magnetic forces of the material's electrons will be affected. This effect is known as Faraday's Law of Magnetic Induction. However, materials can react quite differently to the presence of an external magnetic field. This reaction is dependent on a number of factors, such as the atomic and molecular structure of the material, and the net magnetic field associated with the atoms. The magnetic moments associated with atoms have three origins. These are the electron motion, the change in motion caused by an external magnetic field, and the spin of the electrons.
In most atoms, electrons occur in pairs. Electrons in a pair spin in opposite directions. So, when electrons are paired together, their opposite spins cause their magnetic fields to cancel each other. Therefore, no net magnetic field exists. Alternately, materials with some unpaired electrons will have a net magnetic field and will react more to an external field. Most materials can be classified as diamagnetic, paramagnetic or ferromagnetic.
Diamagnetic:
Diamagnetic materials have a weak, negative susceptibility to magnetic fields. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. Diamagnetic properties arise from the realignment of the electron paths under the influence of an external magnetic field. Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.
If we plot M vs H, we see:
Note that when the field is zero the magnetization is zero. The other characteristic behavior of diamagnetic materials is that the susceptibility is temperature independent. Some well known diamagnetic substances, in units of 10-8 m3/kg, include:
quartz (SiO2) -0.62
Calcite (CaCO3) -0.48
water -0.90
Paramagnetic:
Paramagnetic materials have a small, positive susceptibility to magnetic fields. These materials are slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.
In addition, the efficiency of the field in aligning the moments is opposed by the randomizing effects of temperature. This results in a temperature dependent susceptibility, known as the Curie Law.
At normal temperatures and in moderate fields, the paramagnetic susceptibility is small (but larger than the diamagnetic contribution). Unless the temperature is very low (<<100 K) or the field is very high paramagnetic susceptibility is independent of the applied field. Under these conditions, paramagnetic susceptibility is proportional to the total iron content. Many iron bearing minerals are paramagnetic at room temperature. Some examples, in units of 10-8 m3/kg, include:
Montmorillonite (clay) 13
Nontronite (Fe-rich clay) 65
Biotite (silicate) 79
Siderite(carbonate) 100
Pyrite (sulfide) 30
The paramagnetism of the matrix minerals in natural samples can be significant if the concentration of magnetite is very small. In this case, a paramagnetic correction may be needed.
Ferromagnetic:
Ferromagnetic materials have a large, positive susceptibility to an external magnetic field. They exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. Ferromagnetic materials have some unpaired electrons so their atoms have a net magnetic moment. They get their strong magnetic properties due to the presence of magnetic domains. In these domains, large numbers of atom's moments (1012 to 1015) are aligned parallel so that the magnetic force within the domain is strong. When a ferromagnetic material is in the unmagnitized state, the domains are nearly randomly organized and the net magnetic field for the part as a whole is zero. When a magnetizing force is applied, the domains become aligned to produce a strong magnetic field within the part. Iron, nickel, and cobalt are examples of ferromagnetic materials. Components with these materials are commonly inspected using the magnetic particle method.
Ferromagnetic materials exhibit parallel alignment of moments resulting in large net magnetization even in the absence of a magnetic field. The elements Fe, Ni, and Co and many of their alloys are typical ferromagnetic materials.
Two distinct characteristics of ferromagnetic materials are their
(1) spontaneous magnetization and the existence of
(2) magnetic ordering temperature
Solution:
Repelled due to induction of similar poles.
Solution:
cm is positive at all temperature
Solution:
It will Fall, because diamagnetic substance, moves from stronger magnetic field to weaker field.
1. A square loop OABCO of side l carries a current i. Now, find the magnetic moment of the loop.
Solution:
Magnetic moment of the loop can be written as,
M = i (BC x CO), where the letters in bold denote the vectors
BC = -lk
1. Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i = 2.0 A.
Solution:
By assuming two equal and opposite currents in BE, two current carrying loop (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence
Mnet = √M2 + M2 = √2 M
Where M = ia = (2.0)(0.1)(0.1) = 0.02 A-m2
Mnet = (√2) (0.02) A – m2 = 0.028 A-m2
2. Two identical bar magnets each of length L and pole–strength m are placed at right angles to each other with the north pole of one touching the south pole of the other. Evaluate the magnetic moment of the system.
Solution:
As magnetic moment is a vector, so we know
MR = (M12 + M22 + 2M1M2 cos θ)1/2 with tanΦ M2 sinθ / M1 + M2 cosθ
and as here M1= M2 = mL and θ = 90°, so
MR = (M2 + M2 + 2MM cos 90)1/2 = (√2) mL
And, tanΦ Msin90 / M + Mcos 90 = 1, i.e., Φ = tan–1(1) = 45°