Photoelectric cut-off voltage, Vo = 1.5 V The maximum kinetic energy of the emitted photoelectrons is given as: Ke=eV0 Where, e = Charge on an electron = 1.6 x 10-19 C Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J Therefore, the maximum kinetic energy of the photoelectrons emitted in the given eRead more
Photoelectric cut-off voltage, Vo = 1.5 V
The maximum kinetic energy of the emitted photoelectrons is given as:
Ke=eV0
Where, e = Charge on an electron = 1.6 x 10-19 C
Therefore, Ke = 1.6 x 10-19 x 1.5
= 2.4 x 10-19 J
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 x 10-19J.
Work function of caesium metal, φO = 2.14 eV and frequency of light, v = 6.0 x 1014 Hz Ans (a). The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO Where, h = Planck’s constant = 6.626 x 10⁻34 Js Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140 = 2.485Read more
Work function of caesium metal, φO = 2.14 eV and
frequency of light, v = 6.0 x 1014 Hz
Ans (a).
The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO
Where, h = Planck’s constant = 6.626 x 10⁻34 Js
Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140
= 2.485 -2.140 = 0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
Ans (b).
For stopping potential V0 we can write the equation for kinetic energy as:
K=e V0
Therefore, V0= K/e = (0.345 x 1.6 x10⁻¹⁹) /(1.6 x 10⁻¹⁹ )
Hence ,the stopping potential of the material is 0.345 V
Ans (c).
Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
K = 1/2 x (mv²)
Where, m = Mass of an electron = 9.1 x 10-31 kg
v² =2K /m
= 2 x (0.345 x 1.6 x 10⁻¹⁹ )/(9.1 x 10-31) = 0.1104 x 10¹²
v=3.323 x 10⁵ m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1 Magnitude of earth's magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T Ans (a). The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation: B₁ = μ0 /4π x [M/(R)³] Where, μ0 = PermeabiliRead more
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T
Ans (a).
The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
B₁ = μ0 /4π x [M/(R)³]
Where,
μ0 = Permeability of free space = 4π x 10-7 Tm A-1
When the resultant field is inclined at 45° with earth’s field, B = H
Therefore , μ0 /4π x [M/(R)³] =H = 0.42 x 10⁻4
=> R³ = μ0M /( 0.42 x 10⁻4 x 4π )
= (4π x 10-7 x 5.25 x 10⁻2 )/ ( 0.42 x 10⁻4 x 4π ) = 12.5 x 10⁻⁵
Therefore R = 0.05m = 5 cm
Ans (b).
The magnetic field at a distance R1 from the centre of the magnet on its axis is given as:
B¹ = μ0 /4π x [2M/(R¹)³]
The resultant field is inclined at 45° with earth’s field.
Therefore ,
B¹ =H
μ0 /4π x [2M/(R¹)³] = H
(R¹)³ = μ0 2M/ 4πH
= (4π x 10-7 x 2 x 5.25 x 10⁻2)/ (4π x 0.42 x 10⁻4 )
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as: B₁ = μ0 /4π x [2M/(d₁)³] =H -----------Eq 1 Where, M = Magnetic moment μ0= Permeability of free space, H = Horizontal component of the magnetic field at d₁. If the bar magnet is turned through 180°, then the nRead more
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as:
B₁ = μ0 /4π x [2M/(d₁)³] =H ———–Eq 1
Where, M = Magnetic moment
μ0= Permeability of free space,
H = Horizontal component of the magnetic field at d₁.
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance d₂, on the equatorial line of the magnet can be written as:
B₂ = μ0 /4π x [2M/(d₂)³] =H ———–Eq 2
Equating equations (1) and (2), we get:
2/(d₁)³= 1/(d₂)³
=> (d₂/d₁)³ = 1/2
=> d₂ = d₁ x (1/2)1/3
=> d₂ = 14 x 0.794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
Earth's magnetic field at the given place, H = 0.36 G The magnetic field at a distance d, on the axis of the magnet is given as: B₁ = μ0 /4π x (2M/d³) = H ------------------Eq -1 Where, μ0 = Permeability of free space, M = Magnetic moment The magnetic field at the same distance d, on the equatorialRead more
Earth’s magnetic field at the given place, H = 0.36 G The magnetic field at a distance d, on the axis of the magnet is given as: B₁ = μ0 /4π x (2M/d³) = H ——————Eq -1
Where, μ0 = Permeability of free space, M = Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:
B₂ = μ0 /4π x (M/d³) = H/2 (Using Eq -1)
Total magnetic field ,B = B₁ + B₂
= 0.36 + 0.18 = 0.54 G
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
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.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Photoelectric cut-off voltage, Vo = 1.5 V The maximum kinetic energy of the emitted photoelectrons is given as: Ke=eV0 Where, e = Charge on an electron = 1.6 x 10-19 C Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J Therefore, the maximum kinetic energy of the photoelectrons emitted in the given eRead more
Photoelectric cut-off voltage, Vo = 1.5 V
The maximum kinetic energy of the emitted photoelectrons is given as:
Ke=eV0
Where, e = Charge on an electron = 1.6 x 10-19 C
Therefore, Ke = 1.6 x 10-19 x 1.5
= 2.4 x 10-19 J
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 x 10-19J.
See lessThe work function of caesium metal is 2.14 eV. When light of frequency 6 ×10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?
Work function of caesium metal, φO = 2.14 eV and frequency of light, v = 6.0 x 1014 Hz Ans (a). The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO Where, h = Planck’s constant = 6.626 x 10⁻34 Js Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140 = 2.485Read more
Work function of caesium metal, φO = 2.14 eV and
frequency of light, v = 6.0 x 1014 Hz
Ans (a).
The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO
Where, h = Planck’s constant = 6.626 x 10⁻34 Js
Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140
= 2.485 -2.140 = 0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
Ans (b).
For stopping potential V0 we can write the equation for kinetic energy as:
K=e V0
Therefore, V0= K/e = (0.345 x 1.6 x10⁻¹⁹) /(1.6 x 10⁻¹⁹ )
Hence ,the stopping potential of the material is 0.345 V
Ans (c).
Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
K = 1/2 x (mv²)
Where, m = Mass of an electron = 9.1 x 10-31 kg
v² =2K /m
= 2 x (0.345 x 1.6 x 10⁻¹⁹ )/(9.1 x 10-31) = 0.1104 x 10¹²
v=3.323 x 10⁵ m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
See lessA short bar magnet of magnetic moment 5.25 × 10⁻² J T⁻¹ is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1 Magnitude of earth's magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T Ans (a). The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation: B₁ = μ0 /4π x [M/(R)³] Where, μ0 = PermeabiliRead more
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T
Ans (a).
The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
B₁ = μ0 /4π x [M/(R)³]
Where,
μ0 = Permeability of free space = 4π x 10-7 Tm A-1
When the resultant field is inclined at 45° with earth’s field, B = H
Therefore , μ0 /4π x [M/(R)³] =H = 0.42 x 10⁻4
=> R³ = μ0M /( 0.42 x 10⁻4 x 4π )
= (4π x 10-7 x 5.25 x 10⁻2 )/ ( 0.42 x 10⁻4 x 4π ) = 12.5 x 10⁻⁵
Therefore R = 0.05m = 5 cm
Ans (b).
The magnetic field at a distance R1 from the centre of the magnet on its axis is given as:
B¹ = μ0 /4π x [2M/(R¹)³]
The resultant field is inclined at 45° with earth’s field.
Therefore ,
B¹ =H
μ0 /4π x [2M/(R¹)³] = H
(R¹)³ = μ0 2M/ 4πH
= (4π x 10-7 x 2 x 5.25 x 10⁻2)/ (4π x 0.42 x 10⁻4 )
= 25 x 10⁻⁵
Therefore , R¹ = 0.063 m = 6.3 cm
See lessIf the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as: B₁ = μ0 /4π x [2M/(d₁)³] =H -----------Eq 1 Where, M = Magnetic moment μ0= Permeability of free space, H = Horizontal component of the magnetic field at d₁. If the bar magnet is turned through 180°, then the nRead more
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as:
B₁ = μ0 /4π x [2M/(d₁)³] =H ———–Eq 1
Where, M = Magnetic moment
μ0= Permeability of free space,
H = Horizontal component of the magnetic field at d₁.
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance d₂, on the equatorial line of the magnet can be written as:
B₂ = μ0 /4π x [2M/(d₂)³] =H ———–Eq 2
Equating equations (1) and (2), we get:
2/(d₁)³= 1/(d₂)³
=> (d₂/d₁)³ = 1/2
=> d₂ = d₁ x (1/2)1/3
=> d₂ = 14 x 0.794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
See lessA short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Earth's magnetic field at the given place, H = 0.36 G The magnetic field at a distance d, on the axis of the magnet is given as: B₁ = μ0 /4π x (2M/d³) = H ------------------Eq -1 Where, μ0 = Permeability of free space, M = Magnetic moment The magnetic field at the same distance d, on the equatorialRead more
Earth’s magnetic field at the given place, H = 0.36 G
The magnetic field at a distance d, on the axis of the magnet is given as:
B₁ = μ0 /4π x (2M/d³) = H ——————Eq -1
Where, μ0 = Permeability of free space, M = Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:
B₂ = μ0 /4π x (M/d³) = H/2 (Using Eq -1)
Total magnetic field ,B = B₁ + B₂
= 0.36 + 0.18 = 0.54 G
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
See lessA 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