Magnetic moment of the bar magnet, M = 0.48 J T-1 Ans (a). Distance, d = 10 cm = 0.1 m The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation: B = μ0 /4π x (2M/d³) Where, μ0 = Permeability of free space =4π x 10-7 T m A-1 Therefore ,B = (4π x 10-7 x 2Read more
Magnetic moment of the bar magnet, M = 0.48 J T-1
Ans (a).
Distance, d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:
B = μ0 /4π x (2M/d³)
Where, μ0 = Permeability of free space =4π x 10-7 T m A-1
Therefore ,B = (4π x 10-7 x 2 x 0.48)/[4π x (0.1)³]
The magnetic field is along the S – N direction.
Ans (b).
The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:
Angle of declination, 0 = 12° Angle of dip, δ = 60° Horizontal component of earth's magnetic field, BH = 0.16 G Earth's magnetic field at the given location = B We can relate B and BH as: BH = B cos δ Therefore , B = BH /cos δ = 0.16/cos 60° = 0.32 G Earth's magnetic field lies in the vertical planeRead more
Angle of declination, 0 = 12°
Angle of dip, δ = 60°
Horizontal component of earth’s magnetic field, BH = 0.16 G
Earth’s magnetic field at the given location = B
We can relate B and BH as:
BH = B cos δ
Therefore , B = BH /cos δ = 0.16/cos 60° = 0.32 G
Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.
Horizontal component of earth's magnetic field, BH= 0.35 G Angle made by the needle with the horizontal plane = Angle of dip , δ = 22° Earth's magnetic field strength = B We can relate B and BH as: BH = Bcos0 Therefore B = BH /cosδ 0.35 /cos 22° =0.377 G Hence, the strength of earth's magnetic fielRead more
Horizontal component of earth’s magnetic field,
BH= 0.35 G
Angle made by the needle with the horizontal plane = Angle of dip , δ = 22°
Earth’s magnetic field strength = B We can relate B and BH as:
BH = Bcos0
Therefore B = BH /cosδ
0.35 /cos 22° =0.377 G
Hence, the strength of earth’s magnetic field at the given location is 0.377 G.
Number of turns in the circular coil, N = 16 Radius of the coil, r = 10 cm = 0.1 m Cross-section of the coil, A =πr2 = π x (0.1)2 m2 Current in the coil, I = 0.75 A Magnetic field strength, B = 5.0 x 10-2 T Frequency of oscillations of the coil, v = 2.0 s-1 Therefore , Magnetic moment, M = NIA = NJπRead more
Number of turns in the circular coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Cross-section of the coil, A =πr2 = π x (0.1)2 m2
Current in the coil, I = 0.75 A
Magnetic field strength, B = 5.0 x 10-2 T
Frequency of oscillations of the coil, v = 2.0 s-1
Therefore , Magnetic moment, M = NIA = NJπr2
= 16 x 0.75 x n x (0.1)2 = 0.377 J T-1
Frequency is given by the relation:
v =( 1/2 π) √ (MB/I)
Where,
I = Moment of inertia of the coil
Therefore I = MB /(4π²v²) = (0.377 x 5 x 10⁻²)/( 4π² x 2² )
= 1.19 x 10⁻4 kg m²
Hence, the moment of inertia of the coil about its axis of rotation is 1.19 x 10⁻4 kg m²
Number of turns on the solenoid, n = 2000 Area of cross-section of the solenoid, A = 1.6 x 10-4m2 Current in the solenoid, I = 4 A Ans (a). The magnetic moment along the axis of the solenoid is calculated as: M = nAI = 2000 x 1.6 x 10⁻4 x 4 = 1.28 Am2 Ans (b). Magnetic field, B = 7.5 x 10-2 T AngleRead more
Number of turns on the solenoid, n = 2000
Area of cross-section of the solenoid, A = 1.6 x 10-4m2
Current in the solenoid, I = 4 A
Ans (a).
The magnetic moment along the axis of the solenoid is calculated as:
M = nAI = 2000 x 1.6 x 10⁻4 x 4 = 1.28 Am2
Ans (b).
Magnetic field, B = 7.5 x 10-2 T
Angle between the magnetic field and the axis of the solenoid, 0 = 30°
Torque, τ = MBsinB
= 1.28 x 7.5 x10⁻2 sin 30°
= 4.8 x 10⁻2 Nm
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 x 10⁻2 Nm.
A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Magnetic moment of the bar magnet, M = 0.48 J T-1 Ans (a). Distance, d = 10 cm = 0.1 m The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation: B = μ0 /4π x (2M/d³) Where, μ0 = Permeability of free space =4π x 10-7 T m A-1 Therefore ,B = (4π x 10-7 x 2Read more
Magnetic moment of the bar magnet, M = 0.48 J T-1
Ans (a).
Distance, d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:
B = μ0 /4π x (2M/d³)
Where, μ0 = Permeability of free space =4π x 10-7 T m A-1
Therefore ,B = (4π x 10-7 x 2 x 0.48)/[4π x (0.1)³]
The magnetic field is along the S – N direction.
Ans (b).
The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:
B = μ0 /4π x (M/d³)
=> B = (4π x 10-7 x 0.48)/[4π x (0.1)³]
= 0.48 G
The magnetic field is along the N – S direction.
See lessAt a certain location in Africa, a compass points 12º west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Angle of declination, 0 = 12° Angle of dip, δ = 60° Horizontal component of earth's magnetic field, BH = 0.16 G Earth's magnetic field at the given location = B We can relate B and BH as: BH = B cos δ Therefore , B = BH /cos δ = 0.16/cos 60° = 0.32 G Earth's magnetic field lies in the vertical planeRead more
Angle of declination, 0 = 12°
Angle of dip, δ = 60°
Horizontal component of earth’s magnetic field, BH = 0.16 G
Earth’s magnetic field at the given location = B
We can relate B and BH as:
BH = B cos δ
Therefore , B = BH /cos δ = 0.16/cos 60° = 0.32 G
Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.
See lessA magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22º with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Horizontal component of earth's magnetic field, BH= 0.35 G Angle made by the needle with the horizontal plane = Angle of dip , δ = 22° Earth's magnetic field strength = B We can relate B and BH as: BH = Bcos0 Therefore B = BH /cosδ 0.35 /cos 22° =0.377 G Hence, the strength of earth's magnetic fielRead more
Horizontal component of earth’s magnetic field,
BH= 0.35 G
Angle made by the needle with the horizontal plane = Angle of dip , δ = 22°
Earth’s magnetic field strength = B We can relate B and BH as:
BH = Bcos0
Therefore B = BH /cosδ
0.35 /cos 22° =0.377 G
Hence, the strength of earth’s magnetic field at the given location is 0.377 G.
See lessA circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10⁻² T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s⁻¹. What is the moment of inertia of the coil about its axis of rotation?
Number of turns in the circular coil, N = 16 Radius of the coil, r = 10 cm = 0.1 m Cross-section of the coil, A =πr2 = π x (0.1)2 m2 Current in the coil, I = 0.75 A Magnetic field strength, B = 5.0 x 10-2 T Frequency of oscillations of the coil, v = 2.0 s-1 Therefore , Magnetic moment, M = NIA = NJπRead more
Number of turns in the circular coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Cross-section of the coil, A =πr2 = π x (0.1)2 m2
Current in the coil, I = 0.75 A
Magnetic field strength, B = 5.0 x 10-2 T
Frequency of oscillations of the coil, v = 2.0 s-1
Therefore , Magnetic moment, M = NIA = NJπr2
= 16 x 0.75 x n x (0.1)2 = 0.377 J T-1
Frequency is given by the relation:
v =( 1/2 π) √ (MB/I)
Where,
I = Moment of inertia of the coil
Therefore I = MB /(4π²v²) = (0.377 x 5 x 10⁻²)/( 4π² x 2² )
= 1.19 x 10⁻4 kg m²
Hence, the moment of inertia of the coil about its axis of rotation is 1.19 x 10⁻4 kg m²
See lessA closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m² , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 30º with the axis of the solenoid?
Number of turns on the solenoid, n = 2000 Area of cross-section of the solenoid, A = 1.6 x 10-4m2 Current in the solenoid, I = 4 A Ans (a). The magnetic moment along the axis of the solenoid is calculated as: M = nAI = 2000 x 1.6 x 10⁻4 x 4 = 1.28 Am2 Ans (b). Magnetic field, B = 7.5 x 10-2 T AngleRead more
Number of turns on the solenoid, n = 2000
Area of cross-section of the solenoid, A = 1.6 x 10-4m2
Current in the solenoid, I = 4 A
Ans (a).
The magnetic moment along the axis of the solenoid is calculated as:
M = nAI = 2000 x 1.6 x 10⁻4 x 4 = 1.28 Am2
Ans (b).
Magnetic field, B = 7.5 x 10-2 T
Angle between the magnetic field and the axis of the solenoid, 0 = 30°
Torque, τ = MBsinB
= 1.28 x 7.5 x10⁻2 sin 30°
= 4.8 x 10⁻2 Nm
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 x 10⁻2 Nm.
See less