Number of atomic dipoles, n = 2.0 x 1024 Dipole moment of each atomic dipole, M = 1.5 x 10-23 J T⁻¹ When the magnetic field, Bi = 0.64 T The sample is cooled to a temperature, T₁ = 4.2°K Total dipole moment of the atomic dipole, Mtot = n x M = 2 x 1024 x 1.5 x 10-23 = 30 J T⁻¹ Magnetic saturation isRead more
Number of atomic dipoles, n = 2.0 x 1024
Dipole moment of each atomic dipole, M = 1.5 x 10-23 J T⁻¹
Energy of an electron beam, E = 18 keV = 18 x 103 eV, Charge on an electron, e = 1.6 x 10-19 C, E = 18 x 103 x 1.6 x 10-19 J, Magnetic field, B = 0.04 G, Mass of an electron, me = 9.11 x 10-19 kg Distance up to which the electron beam travels, d = 30 cm = 0.3 m We can write the kinetic energy of theRead more
Energy of an electron beam, E = 18 keV = 18 x 103 eV,
Charge on an electron, e = 1.6 x 10-19 C,
E = 18 x 103 x 1.6 x 10-19 J,
Magnetic field, B = 0.04 G,
Mass of an electron, me = 9.11 x 10-19 kg
Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:
E = 1/2 mv²
v = √ ( 2 x 18 x 103 x 1.6 x 10-19 )/ (9.11 x 10⁻³¹) = 0.795 x 10⁸ m/s
The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.
BeV = mv²/r
Therefore , r = mv /Be
= (9.11 x 10⁻³¹ x 0.795 x 10⁸)/ (0.4 x 10 x 1.6 x 10-19) = 11.3 m
Let the up and down deflection of the electron beam be x = r(1 – cos 0) Where,0 = Angle of declination
sin 0 = d/r
= 0.3 /11.3
0 = sin⁻¹ 0.3/11.3 = 1.521°
And x = 11.3 (1 – cos 1.521° )
= .0039 m = 3.9mm
Therefore ,the up and down deflection of the beam is 3.9mm
Magnitude of one of the magnetic fields, B₁ = 1.2 x 10-2 T Magnitude of the other magnetic field = B₂ Angle between the two fields, 0 = 60° At stable equilibrium, the angle between the dipole and field B₁, 0₁ = 15° Angle between the dipole and field B2, 02 = 0 - 0₁ = 60° - 15° = 45° At rotational eRead more
Magnitude of one of the magnetic fields, B₁ = 1.2 x 10-2 T
Magnitude of the other magnetic field = B₂
Angle between the two fields, 0 = 60°
At stable equilibrium, the angle between the dipole and field B₁, 0₁ = 15°
Angle between the dipole and field B2, 02 = 0 – 0₁ = 60° – 15° = 45°
At rotational equilibrium, the torques between both the fields must balance each other.
.-. Torque due to field B₁ = Torque due to field B2
MB₁ sin0₁ = MB2 sin02
Where,
M = Magnetic moment of the dipole
Therefore , B2 =(B₁ sin0₁ )/sin02
= 1.2 x10⁻2 x sin 15°/sin 45° = 4.39 x 10-3 T
Hence, the magnitude of the other magnetic field is 4.39 x 10-3 T.
Number of turns in the circular coil, N = 30, Radius of the circular coil, r = 12 cm = 0.12 m Current in the coil, I = 0.35 A, Angle of dip, δ = 45° Ans (a). The magnetic field due to current I, at a distance r, is given as: B= μ0 /4π 2πNI/r Where, μ0 = Permeability of free space = 4π x 10⁻7 T m A-1Read more
Number of turns in the circular coil, N = 30,
Radius of the circular coil, r = 12 cm = 0.12 m
Current in the coil, I = 0.35 A, Angle of dip, δ = 45°
Ans (a).
The magnetic field due to current I, at a distance r, is given as:
B= μ0 /4π 2πNI/r
Where, μ0 = Permeability of free space = 4π x 10⁻7 T m A-1 .
B = ( 4π x 10⁻7 x 2π x 30 x 0.35)/ (4π x 0.12) = 5.49 x 10⁻⁵
The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as: BH = B sinδ = 5.49 x 10⁻⁵ sin 45° = 3.88 x 10⁻⁵ T = 0.388 G
Ans (b).
When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 °, the needle will reverse its original direction. In this case, the needle will point from East to West.
Number of horizontal wires in the telephone cable, n = 4 Current in each wire, I = 1.0 A Earth's magnetic field at a location, H = 0.39 G = 0.39 x 10_4T Angle of dip at the location, δ = 35° Angle of declination, 0 ~ 0° For a point 4 cm below the cable: Distance, r = 4 cm = 0.04 m The horizontal comRead more
Number of horizontal wires in the telephone cable, n = 4
Current in each wire, I = 1.0 A
Earth’s magnetic field at a location, H = 0.39 G = 0.39 x 10_4T
Angle of dip at the location, δ = 35°
Angle of declination, 0 ~ 0°
For a point 4 cm below the cable:
Distance, r = 4 cm = 0.04 m
The horizontal component of earth’s magnetic field can be written as:
Hh = H cos δ – B
Where,
B = Magnetic field at 4 cm due to current I in the four wires = 4 x μ0 I/2πR
μ0 = Permeability of free space = 4πx 10-7 Tm A-1
Therefore , B = 4 x (4πx 10-7 x I)/ (2π x .04)
= 0.2 x 10-4 T = 0.2 G
Therefore , Hh = 0.39 cos 35° – 0.2 = 0.39 x 0.819 – 0.2 ≈ 0.12 G
The vertical component of earth’s magnetic field is given as: Hv = H sinδ = 0.39 sin 35° = 0.22 G The angle made by the field with its horizontal component is given as:
0 = tan⁻¹Hv/Hh
= tan⁻¹ 0.22/0.12= 61.39º
The resultant field at the point is given as:
H1 = √ [(H)² + (Hh)²]
=√(0.22)²+ (0.12)² = 0.25 G
For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field: Hh = Hcosδ + B = 0.39 cos 35° + 0.2 = 0.52 G Vertical component of earth’s magnetic field: Hv = Hsinδ = 0.39 sin 35° = 0.22 G
A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles each of dipole moment 1.5 × 10⁻²³ J T⁻¹. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law
Number of atomic dipoles, n = 2.0 x 1024 Dipole moment of each atomic dipole, M = 1.5 x 10-23 J T⁻¹ When the magnetic field, Bi = 0.64 T The sample is cooled to a temperature, T₁ = 4.2°K Total dipole moment of the atomic dipole, Mtot = n x M = 2 x 1024 x 1.5 x 10-23 = 30 J T⁻¹ Magnetic saturation isRead more
Number of atomic dipoles, n = 2.0 x 1024
Dipole moment of each atomic dipole, M = 1.5 x 10-23 J T⁻¹
When the magnetic field, Bi = 0.64 T
The sample is cooled to a temperature, T₁ = 4.2°K
Total dipole moment of the atomic dipole,
Mtot = n x M = 2 x 1024 x 1.5 x 10-23 = 30 J T⁻¹
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, M1 = 15/100 x 30 = 4.5 JT-1
When the magnetic field, B2 = 0.98 T
Temperature, T2 = 2.8°K
Its total dipole moment = M2
According to Curie’s law, we have the ratio of two magnetic dipoles as:
M₂/M₁ = (B₂/B₁ ) x (T₁/T₂)
Therefore, M₂ = (B₂ T₁ M₁)/( B₁T₂)
= (0.98 x 4.2 x 4.5)/(2.8 x 0.64) = 10.336 J T⁻¹
Therefore, 10.336 J T⁻¹ is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.
See lessA monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 × 10⁻¹⁹C). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Energy of an electron beam, E = 18 keV = 18 x 103 eV, Charge on an electron, e = 1.6 x 10-19 C, E = 18 x 103 x 1.6 x 10-19 J, Magnetic field, B = 0.04 G, Mass of an electron, me = 9.11 x 10-19 kg Distance up to which the electron beam travels, d = 30 cm = 0.3 m We can write the kinetic energy of theRead more
Energy of an electron beam, E = 18 keV = 18 x 103 eV,
Charge on an electron, e = 1.6 x 10-19 C,
E = 18 x 103 x 1.6 x 10-19 J,
Magnetic field, B = 0.04 G,
Mass of an electron, me = 9.11 x 10-19 kg
Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:
E = 1/2 mv²
v = √ ( 2 x 18 x 103 x 1.6 x 10-19 )/ (9.11 x 10⁻³¹) = 0.795 x 10⁸ m/s
The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.
BeV = mv²/r
Therefore , r = mv /Be
= (9.11 x 10⁻³¹ x 0.795 x 10⁸)/ (0.4 x 10 x 1.6 x 10-19) = 11.3 m
Let the up and down deflection of the electron beam be x = r(1 – cos 0) Where,0 = Angle of declination
sin 0 = d/r
= 0.3 /11.3
0 = sin⁻¹ 0.3/11.3 = 1.521°
And x = 11.3 (1 – cos 1.521° )
= .0039 m = 3.9mm
Therefore ,the up and down deflection of the beam is 3.9mm
See lessA magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60º, and one of the fields has a magnitude of 1.2 × 10⁻² T. If the dipole comes to stable equilibrium at an angle of 15º with this field, what is the magnitude of the other field?
Magnitude of one of the magnetic fields, B₁ = 1.2 x 10-2 T Magnitude of the other magnetic field = B₂ Angle between the two fields, 0 = 60° At stable equilibrium, the angle between the dipole and field B₁, 0₁ = 15° Angle between the dipole and field B2, 02 = 0 - 0₁ = 60° - 15° = 45° At rotational eRead more
Magnitude of one of the magnetic fields, B₁ = 1.2 x 10-2 T
Magnitude of the other magnetic field = B₂
Angle between the two fields, 0 = 60°
At stable equilibrium, the angle between the dipole and field B₁, 0₁ = 15°
Angle between the dipole and field B2, 02 = 0 – 0₁ = 60° – 15° = 45°
At rotational equilibrium, the torques between both the fields must balance each other.
.-. Torque due to field B₁ = Torque due to field B2
MB₁ sin0₁ = MB2 sin02
Where,
M = Magnetic moment of the dipole
Therefore , B2 =( B₁ sin0₁ )/sin02
= 1.2 x10⁻2 x sin 15°/sin 45° = 4.39 x 10-3 T
Hence, the magnitude of the other magnetic field is 4.39 x 10-3 T.
See lessA compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45º with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. (a) Determine the horizontal component of the earth’s magnetic field at the location. (b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Number of turns in the circular coil, N = 30, Radius of the circular coil, r = 12 cm = 0.12 m Current in the coil, I = 0.35 A, Angle of dip, δ = 45° Ans (a). The magnetic field due to current I, at a distance r, is given as: B= μ0 /4π 2πNI/r Where, μ0 = Permeability of free space = 4π x 10⁻7 T m A-1Read more
Number of turns in the circular coil, N = 30,
Radius of the circular coil, r = 12 cm = 0.12 m
Current in the coil, I = 0.35 A, Angle of dip, δ = 45°
Ans (a).
The magnetic field due to current I, at a distance r, is given as:
B= μ0 /4π 2πNI/r
Where, μ0 = Permeability of free space = 4π x 10⁻7 T m A-1 .
B = ( 4π x 10⁻7 x 2π x 30 x 0.35)/ (4π x 0.12) = 5.49 x 10⁻⁵
The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as: BH = B sinδ = 5.49 x 10⁻⁵ sin 45° = 3.88 x 10⁻⁵ T = 0.388 G
Ans (b).
When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 °, the needle will reverse its original direction. In this case, the needle will point from East to West.
See lessA telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35º. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Number of horizontal wires in the telephone cable, n = 4 Current in each wire, I = 1.0 A Earth's magnetic field at a location, H = 0.39 G = 0.39 x 10_4T Angle of dip at the location, δ = 35° Angle of declination, 0 ~ 0° For a point 4 cm below the cable: Distance, r = 4 cm = 0.04 m The horizontal comRead more
Number of horizontal wires in the telephone cable, n = 4
Current in each wire, I = 1.0 A
Earth’s magnetic field at a location, H = 0.39 G = 0.39 x 10_4T
Angle of dip at the location, δ = 35°
Angle of declination, 0 ~ 0°
For a point 4 cm below the cable:
Distance, r = 4 cm = 0.04 m
The horizontal component of earth’s magnetic field can be written as:
Hh = H cos δ – B
Where,
B = Magnetic field at 4 cm due to current I in the four wires = 4 x μ0 I/2πR
μ0 = Permeability of free space = 4πx 10-7 Tm A-1
Therefore , B = 4 x (4πx 10-7 x I)/ (2π x .04)
= 0.2 x 10-4 T = 0.2 G
Therefore , Hh = 0.39 cos 35° – 0.2 = 0.39 x 0.819 – 0.2 ≈ 0.12 G
The vertical component of earth’s magnetic field is given as: Hv = H sinδ = 0.39 sin 35° = 0.22 G The angle made by the field with its horizontal component is given as:
0 = tan⁻¹Hv/Hh
= tan⁻¹ 0.22/0.12= 61.39º
The resultant field at the point is given as:
H1 = √ [(H)² + (Hh)²]
=√(0.22)²+ (0.12)² = 0.25 G
For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field: Hh = Hcosδ + B = 0.39 cos 35° + 0.2 = 0.52 G Vertical component of earth’s magnetic field: Hv = Hsinδ = 0.39 sin 35° = 0.22 G
Angle, 0 = tan-1 Hv/Hh = tan-1 0.22/0.52= 22.9°
And resultant field:
H2= √ [(H)² + (Hh)²]
= √(0.22)²+ (0.52)2 = 0.56 T
See less