Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2 Current carried by the solenoid changes from 2 A to 4 A. Therefore, change in current in the solenoid, di = 4- 2 = 2A Change iRead more
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid, di = 4- 2 = 2A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as: e =dφ/dt ————Eq-1
Where, φ= Induced flux through the small loop =BA————–Eq -2 B = Magnetic field =μ0ni————–Eq -3
μ0 = Permeability of free space = 4π x10-7 H/m
Hence, equation (i) reduces to:
e =d/dt (BA)
= Aμ0n (di/dt)
= 2 x 10⁻4 x 4π x10-7 x 1500 x 2/ 0.1
= 7.54 x 10⁻⁶ V
Hence, the induced voltage in the loop is 7.54 x 10-6V
Ans (a). As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic flux linked with it increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the antiRead more
Ans (a).
As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic
flux linked with it increases. According to Lenz’s law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the anti-clock wise direction. It means the direction of current will be along adcba.
Ans (b).
As the circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic field linked with it also decreases. By Lenz’s law, the induced current should produce a flux in the direction of original flux. For this, the induced current should flow in the anti-clock wise direction. It means the direction of current will be along a’d’c’b’.
The direction of the induced current in a closed loop is given by Lenz's law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The given pairs of figures show the direction of the induced current when tRead more
The direction of the induced current in a closed loop is given by Lenz’s law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.
Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
Ans (a).The direction of the induced current is along qrpq.
Ans (b).The direction of the induced current is along prqp.
Ans (c).The direction of the induced current is along yzxy.
Ans (d).The direction of the induced current is along zyxz.
Ans (e).The direction of the induced current is along xryx.
Ans (f).No current is induced since the field lines are lying in the plane of the closed loop.
Out of the two relations given, only one is in accordance with classical physics. The magnetic momemtum vector which as a result of orbital Angular Moment is given by , μl=−e/2m l It follows from the definitions ofμl and l. μl=iA=(−e/T)πr2 ...(i) Angular momentum, l=mvr=m(2πr/T)r ...(iiRead more
Out of the two relations given, only one is in accordance with classical physics.
The magnetic momemtum vector which as a result of orbital Angular Moment is given by ,
μl=−e/2m l
It follows from the definitions ofμl and l.
μl=iA=(−e/T)πr2 …(i)
Angular momentum, l=mvr=m(2πr/T)r …(ii)
where r is the radius of the circular orbit, which the electron of mass m and charge (−e) completes in time T.
Divide (i) by (ii), μl/l=[(−e/T)πr2 ]/(m(2πr/T)r )=−e/2m
∴μl=(−e/2m)l
Clearly μl and l will be antiparallel (both being normal to the plane of the orbit)
In contrast, μs/S=e/m. It is obtained on the basis of quantum mechanics.
Mean radius of a Rowland ring, r = 15 cm = 0.15 m Number of turns on a ferromagnetic core, N = 3500 Relative permeability of the core material, μ =800 Magnetising current, I = 1.2 A The magnetic field is given by the relation: B = μrμ0 IN/2πr Where, [μ0 = Permeability of free space = 4π x 10-7 T m ARead more
Mean radius of a Rowland ring, r = 15 cm = 0.15 m
Number of turns on a ferromagnetic core, N = 3500
Relative permeability of the core material, μ =800
Magnetising current, I = 1.2 A
The magnetic field is given by the relation:
B = μrμ0 IN/2πr
Where, [μ0 = Permeability of free space = 4π x 10-7 T m A-1
B = (800 x 4π x 10-7 x 1.2 x 3500) / (2πx 0.15) =4.48 T
Therefore, the magnetic field in the core is 4.48 T.
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 long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2 Current carried by the solenoid changes from 2 A to 4 A. Therefore, change in current in the solenoid, di = 4- 2 = 2A Change iRead more
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid, di = 4- 2 = 2A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as: e =dφ/dt ————Eq-1
Where, φ= Induced flux through the small loop =BA————–Eq -2 B = Magnetic field =μ0ni————–Eq -3
μ0 = Permeability of free space = 4π x10-7 H/m
Hence, equation (i) reduces to:
e =d/dt (BA)
= Aμ0n (di/dt)
= 2 x 10⁻4 x 4π x10-7 x 1500 x 2/ 0.1
= 7.54 x 10⁻⁶ V
Hence, the induced voltage in the loop is 7.54 x 10-6V
See lessUse Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.
Ans (a). As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic flux linked with it increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the antiRead more
Ans (a).
As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic
flux linked with it increases. According to Lenz’s law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the anti-clock wise direction. It means the direction of current will be along adcba.
Ans (b).
As the circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic field linked with it also decreases. By Lenz’s law, the induced current should produce a flux in the direction of original flux. For this, the induced current should flow in the anti-clock wise direction. It means the direction of current will be along a’d’c’b’.
See lessPredict the direction of induced current in the situations described by the following Figures. 6.18(a) to (f).
The direction of the induced current in a closed loop is given by Lenz's law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The given pairs of figures show the direction of the induced current when tRead more
The direction of the induced current in a closed loop is given by Lenz’s law, it states that:
The polarity of induced emf is such that it tends to produce a current
which opposes the change in magnetic flux that produced it.
The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.
Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
Ans (a).The direction of the induced current is along qrpq.
Ans (b).The direction of the induced current is along prqp.
Ans (c).The direction of the induced current is along yzxy.
Ans (d).The direction of the induced current is along zyxz.
Ans (e).The direction of the induced current is along xryx.
Ans (f).No current is induced since the field lines are lying in the plane of the closed loop.
See lessThe magnetic moment vectors µs and µl associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: µs = –(e/m) S, µl = –(e/2m)l Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Out of the two relations given, only one is in accordance with classical physics. The magnetic momemtum vector which as a result of orbital Angular Moment is given by , μl=−e/2m l It follows from the definitions ofμl and l. μl=iA=(−e/T)πr2 ...(i) Angular momentum, l=mvr=m(2πr/T)r ...(iiRead more
Out of the two relations given, only one is in accordance with classical physics.
The magnetic momemtum vector which as a result of orbital Angular Moment is given by ,
μl=−e/2m l
It follows from the definitions ofμl and l.
μl=iA=(−e/T)πr2 …(i)
Angular momentum, l=mvr=m(2πr/T)r …(ii)
where r is the radius of the circular orbit, which the electron of mass m and charge (−e) completes in time T.
Divide (i) by (ii), μl/l=[(−e/T)πr2 ]/(m(2πr/T)r )=−e/2m
∴μl=(−e/2m)l
Clearly μl and l will be antiparallel (both being normal to the plane of the orbit)
In contrast, μs/S=e/m. It is obtained on the basis of quantum mechanics.
See lessA Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Mean radius of a Rowland ring, r = 15 cm = 0.15 m Number of turns on a ferromagnetic core, N = 3500 Relative permeability of the core material, μ =800 Magnetising current, I = 1.2 A The magnetic field is given by the relation: B = μrμ0 IN/2πr Where, [μ0 = Permeability of free space = 4π x 10-7 T m ARead more
Mean radius of a Rowland ring, r = 15 cm = 0.15 m
Number of turns on a ferromagnetic core, N = 3500
Relative permeability of the core material, μ =800
Magnetising current, I = 1.2 A
The magnetic field is given by the relation:
B = μrμ0 IN/2πr
Where, [μ0 = Permeability of free space = 4π x 10-7 T m A-1
B = (800 x 4π x 10-7 x 1.2 x 3500) / (2πx 0.15) =4.48 T
Therefore, the magnetic field in the core is 4.48 T.
See lessA 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