Ans (a). Earth's magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth's magnetic field with the time cannot be neglected. Ans (b). Earth's core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considRead more
Ans (a).
Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected.
Ans (b).
Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth’s magnetism.
Ans (c).
The radioactivity in earth’s interior is the source of energy that sustains the currents in the outer conducting regions of earth’s core. These charged currents are considered to be responsible for earth’s magnetism.
Ans (d).
Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.
Ans (e).
Earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.
Ans (f).
An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.
Ans (a). The three independent quantities conventionally used for specifying earth's magnetic field are: Magnetic declination, Angle of dip, and Horizontal component of earth's magnetic field. Ans (b). The angle of dip at a point depends on how far the point is located with respect to the North PoleRead more
Ans (a).
The three independent quantities conventionally used for specifying earth’s magnetic field are:
Magnetic declination,
Angle of dip, and
Horizontal component of earth’s magnetic field.
Ans (b).
The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.
Ans (c).
It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.
Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.
Ans (d).
If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.
Ans (e).
Magnetic moment, M = 8 x 1022 J T⁻1
Radius of earth, r = 6.4 x 106 m
Magnetic field strength , B = (μ0 /4π) x ( M/r³)
Where, μ0 = Permeability of free space = 4π x 10⁻7 T m A-1
Therefore , B= (4π x 10⁻7 x 8 x 1022)/ (4π x 6.4 x 106) =0.3 G
This quantity is of the order of magnitude of the observed field on earth.
Ans (f).
Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.
Resistance of the galvanometer coil, G = 15 Ω Current for which the galvanometer shows full scale deflection, Ig= 4mA = 4 x 10⁻³ A Range of the ammeter is 0, which needs to be converted to 6 A. Therefore ,Current,I = 6 A A shunt resistor of resistance S is to be connected in parallel with the galvanRead more
Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
Ig= 4mA = 4 x 10⁻³ A
Range of the ammeter is 0, which needs to be converted to 6 A.
Therefore ,Current,I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
S =(Ig G)/(I -Ig) = (4 x 10⁻³ x 15)/(6 – 4 x 10⁻³)
=(6 x 10⁻² )/(6 – 0.004) = 0.06 /5.996
≈ 0.01 Ω = 10 mΩ
Hence, a shunt resistor is to be connected in parallel with the galvanometer.
Resistance of the galvanometer coil, G = 12 Ω Current for which there is full scale deflection, Ig = 3 mA = 3 x 10⁻3 A Range of the voltmeter is 0, which needs to be converted to 18 V. Therefore ,V = 18 V Let a resistor of resistance R be connected in series with the galvanometer to convert it intoRead more
Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection, Ig = 3 mA = 3 x 10⁻3 A Range of the voltmeter is 0, which needs to be converted to 18 V.
Therefore ,V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
R = V/Ig – G
= [18 /(3 x 10⁻3) ]-12 = 6000 -12 = 5988 Ω
Hence, a resistor of resistance 5988 Ω is to be connected in series with the galvanometer.
Length of the solenoid, L = 60 cm = 0.6 m Radius of the solenoid, r = 4.0 cm = 0.04 m It is given that there are 3 layers of windings of 300 turns each. Total number of turns, n = 3 x 300 = 900, Length of the wire, 1 = 2 cm = 0.02 m Mass of the wire, m = 2.5 g = 2.5 x 10⁻3 kg Current flowing throughRead more
Length of the solenoid, L = 60 cm = 0.6 m
Radius of the solenoid, r = 4.0 cm = 0.04 m
It is given that there are 3 layers of windings of 300 turns each.
Total number of turns, n = 3 x 300 = 900,
Length of the wire, 1 = 2 cm = 0.02 m
Mass of the wire, m = 2.5 g = 2.5 x 10⁻3 kg
Current flowing through the wire, i = 6 A
Acceleration due to gravity, g = 9.8 m/s2
Magnetic field produced inside the solenoid,
B= (μ0nI)/L
Where, μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
I = Current flowing through the windings of the solenoid
Magnetic force is given by the relation,
F = Bil
= (μ0nI)/L x il
Also, the force on the wire is equal to the weight of the wire,
Therefore , mg=(μ0nI)/L x il
I = (mgL)/(μ0nil) = (2.5 x 10⁻3 x 9.8 x 0.6 )/ ( 4π x 10⁻⁷ x 900 x 0.02 x 6)
=108 A
Hence, the current flowing through the solenoid is 108 A.
Answer the following questions: (a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably? (b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why? (c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents? (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past? (e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion? (f) Interstellar space has an extremely weak magnetic field of the order of 10⁻¹² T. Can such a weak field be of any significant consequence? Explain. [Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]
Ans (a). Earth's magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth's magnetic field with the time cannot be neglected. Ans (b). Earth's core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considRead more
Ans (a).
Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected.
Ans (b).
Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth’s magnetism.
Ans (c).
The radioactivity in earth’s interior is the source of energy that sustains the currents in the outer conducting regions of earth’s core. These charged currents are considered to be responsible for earth’s magnetism.
Ans (d).
Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.
Ans (e).
Earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.
Ans (f).
An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.
See lessAnswer the following questions regarding earth’s magnetism: (a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field. (b) The angle of dip at a location in southern India is about 18º. Would you expect a greater or smaller dip angle in Britain? (c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground? (d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole? (e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 10²² J T⁻¹ located at its centre. Check the order of magnitude of this number in some way. (f ) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
Ans (a). The three independent quantities conventionally used for specifying earth's magnetic field are: Magnetic declination, Angle of dip, and Horizontal component of earth's magnetic field. Ans (b). The angle of dip at a point depends on how far the point is located with respect to the North PoleRead more
Ans (a).
The three independent quantities conventionally used for specifying earth’s magnetic field are:
Ans (b).
The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.
Ans (c).
It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.
Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.
Ans (d).
If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.
Ans (e).
Magnetic moment, M = 8 x 1022 J T⁻1
Radius of earth, r = 6.4 x 106 m
Magnetic field strength , B = (μ0 /4π) x ( M/r³)
Where, μ0 = Permeability of free space = 4π x 10⁻7 T m A-1
Therefore , B= (4π x 10⁻7 x 8 x 1022)/ (4π x 6.4 x 106) =0.3 G
This quantity is of the order of magnitude of the observed field on earth.
Ans (f).
Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.
See lessA galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
Resistance of the galvanometer coil, G = 15 Ω Current for which the galvanometer shows full scale deflection, Ig= 4mA = 4 x 10⁻³ A Range of the ammeter is 0, which needs to be converted to 6 A. Therefore ,Current,I = 6 A A shunt resistor of resistance S is to be connected in parallel with the galvanRead more
Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
Ig= 4mA = 4 x 10⁻³ A
Range of the ammeter is 0, which needs to be converted to 6 A.
Therefore ,Current,I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
S =(Ig G)/(I -Ig) = (4 x 10⁻³ x 15)/(6 – 4 x 10⁻³)
=(6 x 10⁻² )/(6 – 0.004) = 0.06 /5.996
≈ 0.01 Ω = 10 mΩ
Hence, a shunt resistor is to be connected in parallel with the galvanometer.
See lessA galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
Resistance of the galvanometer coil, G = 12 Ω Current for which there is full scale deflection, Ig = 3 mA = 3 x 10⁻3 A Range of the voltmeter is 0, which needs to be converted to 18 V. Therefore ,V = 18 V Let a resistor of resistance R be connected in series with the galvanometer to convert it intoRead more
Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection, Ig = 3 mA = 3 x 10⁻3 A Range of the voltmeter is 0, which needs to be converted to 18 V.
Therefore ,V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
R = V/Ig – G
= [18 /(3 x 10⁻3) ]-12 = 6000 -12 = 5988 Ω
Hence, a resistor of resistance 5988 Ω is to be connected in series with the galvanometer.
See lessA solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s⁻².
Length of the solenoid, L = 60 cm = 0.6 m Radius of the solenoid, r = 4.0 cm = 0.04 m It is given that there are 3 layers of windings of 300 turns each. Total number of turns, n = 3 x 300 = 900, Length of the wire, 1 = 2 cm = 0.02 m Mass of the wire, m = 2.5 g = 2.5 x 10⁻3 kg Current flowing throughRead more
Length of the solenoid, L = 60 cm = 0.6 m
Radius of the solenoid, r = 4.0 cm = 0.04 m
It is given that there are 3 layers of windings of 300 turns each.
Total number of turns, n = 3 x 300 = 900,
Length of the wire, 1 = 2 cm = 0.02 m
Mass of the wire, m = 2.5 g = 2.5 x 10⁻3 kg
Current flowing through the wire, i = 6 A
Acceleration due to gravity, g = 9.8 m/s2
Magnetic field produced inside the solenoid,
B= (μ0nI)/L
Where, μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
I = Current flowing through the windings of the solenoid
Magnetic force is given by the relation,
F = Bil
= (μ0nI)/L x il
Also, the force on the wire is equal to the weight of the wire,
Therefore , mg=(μ0nI)/L x il
I = (mgL)/(μ0nil) = (2.5 x 10⁻3 x 9.8 x 0.6 )/ ( 4π x 10⁻⁷ x 900 x 0.02 x 6)
=108 A
Hence, the current flowing through the solenoid is 108 A.
See less