Magnetic field, B = 0.75 T Accelerating voltage, V = 15 kV = 15 x 103 V Electrostatic field, E = 9 x 105 V m⁻¹ Mass of the electron = m, Charge of the electron = e, Velocity of the electron = v Kinetic energy of the electron = eV =>1/2 x mv2 =eV Therefore e/m = v²/2V---------------------- Eq-1 SiRead more
Magnetic field, B = 0.75 T
Accelerating voltage, V = 15 kV = 15 x 103 V
Electrostatic field, E = 9 x 105 V m⁻¹
Mass of the electron = m,
Charge of the electron = e,
Velocity of the electron = v
Kinetic energy of the electron = eV
=>1/2 x mv2 =eV
Therefore e/m = v²/2V———————- Eq-1
Since the particle remains un-deflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
Therefore eE =evB
v = E/B—————————–Eq-2
Putting Eq-2 in equation 1, we get
e/m = 1/2 x (E/B)² /V = E²/2VB²
= (9 x 105 )²/[2 x 15000 x (0.75)²]
=4.8 x 10⁷ C/Kg
This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He++ Li++, etc
Magnetic field strength, B = 0.15 T Charge on the electron, e = 1.6 x 10⁻¹9C Mass of the electron, m = 9.1 x 10-31 kg Potential difference, V = 2.0 kV = 2 x 103 V Thus, kinetic energy of the electron = eV => eV = 1/2 x mv² v = √ (2eV/m)-----------------------Eq-1 Where, v = velocity of the electrRead more
Magnetic field strength, B = 0.15 T
Charge on the electron, e = 1.6 x 10⁻¹9C
Mass of the electron, m = 9.1 x 10-31 kg
Potential difference, V = 2.0 kV = 2 x 103 V
Thus, kinetic energy of the electron = eV
=> eV = 1/2 x mv²
v = √ (2eV/m)———————–Eq-1
Where, v = velocity of the electron
Ans (a).
Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.
Magnetic force on the electron is given by the relation, Bev
Centripetal force mv²/r
Therefore Bev = mv²/r
r = mv/Be———————–Eq-2
From equations 1 and 2 ,we get
r = m/Be [2eV/m]½
= (9.1 x 10-31 )/ (0.15 x1.6 x 10⁻¹9 ) x [(2 x 1.6 x 10⁻¹9 x 2 x 103)/9.1 x 10-31 ]½
= 100.55 x 10⁻⁵
= 1.01 x 10⁻³m
= 1 mm
Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
Ans (b).
When the field makes an angle 0 of 30° with initial velocity, the initial velocity will be,
v₁= vsin0
From equation (2), we can write the expression for new radius as:
r₁ =mv₁/Be = mvsin0/Be
= (9.1 x 10-31 )/ (0.15 x1.6 x 10⁻¹9 ) x [(2 x 1.6 x 10⁻¹9 x 2 x 103)/9.1 x 10-31 ]½ x sin 30º
= 0.5 x 10-3 m
= 0.5mm
Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
Ans (a). The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field. Ans (b). Yes, the final speed of the charged particle will be equal to its initial speed. This is because maRead more
Ans (a).
The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.
Ans (b).
Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.
Ans (c).
An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain un-deflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.
Inner radius of the toroid, r₁ = 25 cm = 0.25 m Outer radius of the toroid, r₂ = 26 cm = 0.26 m Number of turns on the coil, N = 3500 Current in the coil, I = 11 A Ans (a). Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core oRead more
Inner radius of the toroid, r₁ = 25 cm = 0.25 m
Outer radius of the toroid, r₂ = 26 cm = 0.26 m
Number of turns on the coil, N = 3500
Current in the coil, I = 11 A
Ans (a).
Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core of a toroid is given by the relation, B =
Where, μ0 = Permeability of free space = 4π x 10⁻⁷ TmA⁻¹
l = length of toroid = 2π (r₁ + r₂)/2=2 π (0.25 +0.26)=0.51 π
Therefore B = (4π x 10⁻⁷ x 3500 x 11 )/0.51 π = 3 .0 x 10⁻² T
Ans (c).
Magnetic field in the empty space surrounded by the toroid is zero.
Magnetic field strength, B = 100 G = 100 x 10⁻⁴ T Number of turns per unit length, n = 1000 turns m⁻1 Current flowing in the coil, I = 15 A Permeability of free space, μ0 = 4π x 10-7 T m A-1 Magnetic field is given by the relation, B = μ0 nl =>nl =B/μ0 = (100 x 10⁻⁴ )/(4π x 10-7 ) = 7957.74 ≈ 80Read more
Magnetic field strength, B = 100 G = 100 x 10⁻⁴ T
Number of turns per unit length, n = 1000 turns m⁻1
Current flowing in the coil, I = 15 A
Permeability of free space, μ0 = 4π x 10-7 T m A-1
Magnetic field is given by the relation,
B = μ0 nl
=>nl =B/μ0 = (100 x 10⁻⁴ )/(4π x 10-7 )
= 7957.74 ≈ 8000 Am⁻¹
If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.
A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10⁻⁵ V m⁻¹, make a simple guess as to what the beam contains. Why is the answer not unique?
Magnetic field, B = 0.75 T Accelerating voltage, V = 15 kV = 15 x 103 V Electrostatic field, E = 9 x 105 V m⁻¹ Mass of the electron = m, Charge of the electron = e, Velocity of the electron = v Kinetic energy of the electron = eV =>1/2 x mv2 =eV Therefore e/m = v²/2V---------------------- Eq-1 SiRead more
Magnetic field, B = 0.75 T
Accelerating voltage, V = 15 kV = 15 x 103 V
Electrostatic field, E = 9 x 105 V m⁻¹
Mass of the electron = m,
Charge of the electron = e,
Velocity of the electron = v
Kinetic energy of the electron = eV
=>1/2 x mv2 =eV
Therefore e/m = v²/2V———————- Eq-1
Since the particle remains un-deflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
Therefore eE =evB
v = E/B—————————–Eq-2
Putting Eq-2 in equation 1, we get
e/m = 1/2 x (E/B)² /V = E²/2VB²
= (9 x 105 )²/[2 x 15000 x (0.75)²]
=4.8 x 10⁷ C/Kg
This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He++ Li++, etc
See lessAn electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.
Magnetic field strength, B = 0.15 T Charge on the electron, e = 1.6 x 10⁻¹9C Mass of the electron, m = 9.1 x 10-31 kg Potential difference, V = 2.0 kV = 2 x 103 V Thus, kinetic energy of the electron = eV => eV = 1/2 x mv² v = √ (2eV/m)-----------------------Eq-1 Where, v = velocity of the electrRead more
Magnetic field strength, B = 0.15 T
Charge on the electron, e = 1.6 x 10⁻¹9C
Mass of the electron, m = 9.1 x 10-31 kg
Potential difference, V = 2.0 kV = 2 x 103 V
Thus, kinetic energy of the electron = eV
=> eV = 1/2 x mv²
v = √ (2eV/m)———————–Eq-1
Where, v = velocity of the electron
Ans (a).
Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.
Magnetic force on the electron is given by the relation, Bev
Centripetal force mv²/r
Therefore Bev = mv²/r
r = mv/Be———————–Eq-2
From equations 1 and 2 ,we get
r = m/Be [2eV/m]½
= (9.1 x 10-31 )/ (0.15 x1.6 x 10⁻¹9 ) x [(2 x 1.6 x 10⁻¹9 x 2 x 103)/9.1 x 10-31 ]½
= 100.55 x 10⁻⁵
= 1.01 x 10⁻³m
= 1 mm
Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
Ans (b).
When the field makes an angle 0 of 30° with initial velocity, the initial velocity will be,
v₁= vsin0
From equation (2), we can write the expression for new radius as:
r₁ =mv₁/Be = mvsin0/Be
= (9.1 x 10-31 )/ (0.15 x1.6 x 10⁻¹9 ) x [(2 x 1.6 x 10⁻¹9 x 2 x 103)/9.1 x 10-31 ]½ x sin 30º
= 0.5 x 10-3 m
= 0.5mm
Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
See lessAnswer the following questions: (a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle? (b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment? (c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Ans (a). The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field. Ans (b). Yes, the final speed of the charged particle will be equal to its initial speed. This is because maRead more
Ans (a).
The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.
Ans (b).
Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.
Ans (c).
An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain un-deflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.
See lessA toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.
Inner radius of the toroid, r₁ = 25 cm = 0.25 m Outer radius of the toroid, r₂ = 26 cm = 0.26 m Number of turns on the coil, N = 3500 Current in the coil, I = 11 A Ans (a). Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core oRead more
Inner radius of the toroid, r₁ = 25 cm = 0.25 m
Outer radius of the toroid, r₂ = 26 cm = 0.26 m
Number of turns on the coil, N = 3500
Current in the coil, I = 11 A
Ans (a).
Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core of a toroid is given by the relation, B =
Where, μ0 = Permeability of free space = 4π x 10⁻⁷ TmA⁻¹
l = length of toroid = 2π (r₁ + r₂)/2=2 π (0.25 +0.26)=0.51 π
Therefore B = (4π x 10⁻⁷ x 3500 x 11 )/0.51 π = 3 .0 x 10⁻² T
Ans (c).
Magnetic field in the empty space surrounded by the toroid is zero.
See lessA magnetic field of 100 G (1 G = 10⁻⁴ T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10⁻³ m² . The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m⁻¹. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Magnetic field strength, B = 100 G = 100 x 10⁻⁴ T Number of turns per unit length, n = 1000 turns m⁻1 Current flowing in the coil, I = 15 A Permeability of free space, μ0 = 4π x 10-7 T m A-1 Magnetic field is given by the relation, B = μ0 nl =>nl =B/μ0 = (100 x 10⁻⁴ )/(4π x 10-7 ) = 7957.74 ≈ 80Read more
Magnetic field strength, B = 100 G = 100 x 10⁻⁴ T
Number of turns per unit length, n = 1000 turns m⁻1
Current flowing in the coil, I = 15 A
Permeability of free space, μ0 = 4π x 10-7 T m A-1
Magnetic field is given by the relation,
B = μ0 nl
=>nl =B/μ0 = (100 x 10⁻⁴ )/(4π x 10-7 )
= 7957.74 ≈ 8000 Am⁻¹
If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.
See less