Number of turns on the circular coil, n = 20 Radius of the coil, r = 10 cm = 0.1 m Magnetic field strength, B = 0.10 T Current in the coil, 1 = 5.0 A Ans (a). The total torque on the coil is zero because the field is uniform. Ans (b). The total force on the coil is zero because the field is uniform.Read more
Number of turns on the circular coil, n = 20
Radius of the coil, r = 10 cm = 0.1 m
Magnetic field strength, B = 0.10 T
Current in the coil, 1 = 5.0 A
Ans (a).
The total torque on the coil is zero because the field is uniform.
Ans (b).
The total force on the coil is zero because the field is uniform.
Ans (c).
Cross-sectional area of copper coil, A = 10-5 m2
Number of free electrons per cubic meter in copper, N = 1029/m3
Charge on the electron, e = 1.6 x 10-19 C
Magnetic force, F = Bevd Where, Vd = Drift velocity of electrons
= I/NeA
Therefore ,
F =BeI/NeA
=(0.10 x 5.0)/(1029 x 10-5 )N
Hence, the average force on each electron is 5 x 10–25N.
Magnetic field strength, B = 3000 G = 3000 x 10-4 T = 0.3 T Length of the rectangular loop, l=10 cm Width of the rectangular loop, b = 5 cm Area of the loop, A = l x b = 10x5 = 50 cm2 = 50 x 10⁻⁴m² Current in the loop, I = 12 A Now, taking the anti-clockwise direction of the current as positive andRead more
Magnetic field strength, B = 3000 G = 3000 x 10-4 T = 0.3 T
Length of the rectangular loop, l=10 cm
Width of the rectangular loop, b = 5 cm
Area of the loop, A = l x b = 10×5 = 50 cm2 = 50 x 10⁻⁴m²
Current in the loop, I = 12 A
Now, taking the anti-clockwise direction of the current as positive and vice-versa:
Ans (a).
Torque, τ= IA x B (τ,A & B in vector forms)
From the given figure, it can be observed that A is normal to the y-z plane and B is directed along the z-axis.
Therefore,
τ = 12 x (50 x 10⁻⁴ ) i x 0.3k (i ,j & k in vector forms)
= -1.8×10⁻2j Nm
The torque is 1.8×102 N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.
Ans (b).
This case is similar to case (a). Hence, the answer is the same as (a).
Ans (c).
Torque, τ= IAxB (τ,A & B in vector forms)
From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.
Therefore ,
τ= -12 x (50 x 10⁻⁴ ) i x 0.3k (i ,j & k are vectors)
= —1.8×10-2 i Nm
The torque is 1.8×10-2 Nm along the negative x direction and the force is zero.
Ans (d).
Magnitude of torque is given as:
|τ| = IAB
= 12x50x10⁻4x0.3 = 1.8×10⁻2Nm
The torque is 1.8×10⁻2 N m at an angle of 240° with positive x direction. The force is zero.
Ans (e).
Torque, τ = IAxB = (50×10⁻4 x 12) k x 0.3k = 0 (τ,A & B in vector forms) and (i ,j & k are vectors)
Hence, the torque is zero. The force is also zero.
Ans (f).
Torque τ = IAxB (τ,A & B in vector forms) and (i ,j & k are vectors)
= (50×10⁻4x 12 ) k x 0.3 k = 0
Hence, the torque is zero. The force is also zero.
In case (e), the direction of IA and B is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.
Whereas, in case (f), the direction of IA and B is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.
Magnetic field strength, B = 1.5 T Radius of the cylindrical region, r = 10 cm = 0.1 m Current in the wire passing through the cylindrical region,I = 7 A Ans (a). If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus, l = 2r = 0.2 m Angle betweeRead more
Magnetic field strength, B = 1.5 T
Radius of the cylindrical region, r = 10 cm = 0.1 m
Current in the wire passing through the cylindrical region,I = 7 A
Ans (a).
If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical
region. Thus, l = 2r = 0.2 m
Angle between magnetic field and current, 0 = 90°
Magnetic force acting on the wire is given by the relation,
F = BIl sin 0 = 1.5 x 7 x 0.2 x sin 90° = 2.1 N
Hence, a force of 2.1 N acts on the wire in a vertically downward direction.
Ans (b).
New length of the wire after turning it to the Northeast-Northwest direction can be given as:
l₁ = l sin0
Angle between magnetic field and current, 0 = 45°
Force on the wire,
F = BIl₁sin 0 = Bll = 1.5×7 x 0.2 = 2.1 N
Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle 0 because / sin0 is fixed.
Ans (c).
The wire is lowered from the axis by distance, d = 6.0 cm
Let l₂ be the new length of the wire.
Therefore (l₂/2)² = 4 (10 + 6) = 4 (16)
Therefore l₂ =8 x 2 =16 cm =0.16m
Magnetic force exerted on the wire,
F₂= BIl2
= 1.5 x 7 x 0.16 = 1.68 N
Hence, a force of 1.68 N acts in a vertically downward direction on the wire
Current in both wires, I = 300 A Distance between the wires, r = 1.5 cm = 0.015 m Length of the two wires, l = 70 cm = 0.7 m Force between the two wires is given by the relation, F = μ0 I²/2πr Where, μ0= Permeability of free space = 4π x 10⁻⁷ Tm A⁻1 F = 4π x 10⁻⁷ x (300 )² / (2 π x 0.015) = 1.2 N/mRead more
Current in both wires, I = 300 A
Distance between the wires, r = 1.5 cm = 0.015 m
Length of the two wires, l = 70 cm = 0.7 m
Force between the two wires is given by the relation,
F = μ0 I²/2πr
Where,
μ0= Permeability of free space = 4π x 10⁻⁷ Tm A⁻1
F = 4π x 10⁻⁷ x (300 )² / (2 π x 0.015) = 1.2 N/m
Since the direction of the current in the wires is opposite, a repulsive force exists between them.
Length of the rod, 1 = 0.45 m Mass suspended by the wires, m = 60 g = 60 x 10-3 kg Acceleration due to gravity, g = 9.8 m/s2, Current in the rod flowing through the wire, I = 5A Ans (a). Magnetic field (B) is equal and opposite to the weight of the wire i.e., BIl=mg :.B =mg/Il = (60 x 10-3 x 9.8)/Read more
Length of the rod, 1 = 0.45 m
Mass suspended by the wires, m = 60 g = 60 x 10-3 kg
Acceleration due to gravity, g = 9.8 m/s2, Current in the rod flowing through the wire, I = 5A
Ans (a).
Magnetic field (B) is equal and opposite to the weight of the wire i.e.,
BIl=mg
:.B =mg/Il = (60 x 10-3 x 9.8)/ (5 x 0.45) =0.26 T
A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up in order to get zero tension in the wire. The magnetic field should be such that Fleming’s left hand rule gives an upward magnetic force.
Ans (b).
If the direction of the current is revered, then the force due to magnetic field and the weight of the wire acts in a vertically downward direction.
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 circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the (a) total torque on the coil, (b) total force on the coil, (c) average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area 10⁻⁵ m² , and the free electron density in copper is given to be about 10²⁹m⁻³.)
Number of turns on the circular coil, n = 20 Radius of the coil, r = 10 cm = 0.1 m Magnetic field strength, B = 0.10 T Current in the coil, 1 = 5.0 A Ans (a). The total torque on the coil is zero because the field is uniform. Ans (b). The total force on the coil is zero because the field is uniform.Read more
Number of turns on the circular coil, n = 20
Radius of the coil, r = 10 cm = 0.1 m
Magnetic field strength, B = 0.10 T
Current in the coil, 1 = 5.0 A
Ans (a).
The total torque on the coil is zero because the field is uniform.
Ans (b).
The total force on the coil is zero because the field is uniform.
Ans (c).
Cross-sectional area of copper coil, A = 10-5 m2
Number of free electrons per cubic meter in copper, N = 1029/m3
Charge on the electron, e = 1.6 x 10-19 C
Magnetic force, F = Bevd Where, Vd = Drift velocity of electrons
= I/NeA
Therefore ,
F =BeI/NeA
=(0.10 x 5.0)/(1029 x 10-5 )N
Hence, the average force on each electron is 5 x 10–25N.
See lessA uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?
Magnetic field strength, B = 3000 G = 3000 x 10-4 T = 0.3 T Length of the rectangular loop, l=10 cm Width of the rectangular loop, b = 5 cm Area of the loop, A = l x b = 10x5 = 50 cm2 = 50 x 10⁻⁴m² Current in the loop, I = 12 A Now, taking the anti-clockwise direction of the current as positive andRead more
Magnetic field strength, B = 3000 G = 3000 x 10-4 T = 0.3 T
Length of the rectangular loop, l=10 cm
Width of the rectangular loop, b = 5 cm
Area of the loop, A = l x b = 10×5 = 50 cm2 = 50 x 10⁻⁴m²
Current in the loop, I = 12 A
Now, taking the anti-clockwise direction of the current as positive and vice-versa:
Ans (a).
Torque, τ= IA x B (τ,A & B in vector forms)
From the given figure, it can be observed that A is normal to the y-z plane and B is directed along the z-axis.
Therefore,
τ = 12 x (50 x 10⁻⁴ ) i x 0.3k (i ,j & k in vector forms)
= -1.8×10⁻2j Nm
The torque is 1.8×102 N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.
Ans (b).
This case is similar to case (a). Hence, the answer is the same as (a).
Ans (c).
Torque, τ= IAxB (τ,A & B in vector forms)
From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.
Therefore ,
τ= -12 x (50 x 10⁻⁴ ) i x 0.3k (i ,j & k are vectors)
= —1.8×10-2 i Nm
The torque is 1.8×10-2 Nm along the negative x direction and the force is zero.
Ans (d).
Magnitude of torque is given as:
|τ| = IAB
= 12x50x10⁻4x0.3 = 1.8×10⁻2Nm
The torque is 1.8×10⁻2 N m at an angle of 240° with positive x direction. The force is zero.
Ans (e).
Torque, τ = IAxB = (50×10⁻4 x 12) k x 0.3k = 0 (τ,A & B in vector forms) and (i ,j & k are vectors)
Hence, the torque is zero. The force is also zero.
Ans (f).
Torque τ = IAxB (τ,A & B in vector forms) and (i ,j & k are vectors)
= (50×10⁻4x 12 ) k x 0.3 k = 0
Hence, the torque is zero. The force is also zero.
In case (e), the direction of IA and B is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.
Whereas, in case (f), the direction of IA and B is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.
See lessA uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if, (a) the wire intersects the axis, (b) the wire is turned from N-S to northeast-northwest direction, (c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Magnetic field strength, B = 1.5 T Radius of the cylindrical region, r = 10 cm = 0.1 m Current in the wire passing through the cylindrical region,I = 7 A Ans (a). If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus, l = 2r = 0.2 m Angle betweeRead more
Magnetic field strength, B = 1.5 T
Radius of the cylindrical region, r = 10 cm = 0.1 m
Current in the wire passing through the cylindrical region,I = 7 A
Ans (a).
If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical
region. Thus, l = 2r = 0.2 m
Angle between magnetic field and current, 0 = 90°
Magnetic force acting on the wire is given by the relation,
F = BIl sin 0 = 1.5 x 7 x 0.2 x sin 90° = 2.1 N
Hence, a force of 2.1 N acts on the wire in a vertically downward direction.
Ans (b).
New length of the wire after turning it to the Northeast-Northwest direction can be given as:
l₁ = l sin0
Angle between magnetic field and current, 0 = 45°
Force on the wire,
F = BIl₁sin 0 = Bll = 1.5×7 x 0.2 = 2.1 N
Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle 0 because / sin0 is fixed.
Ans (c).
The wire is lowered from the axis by distance, d = 6.0 cm
Let l₂ be the new length of the wire.
Therefore (l₂/2)² = 4 (10 + 6) = 4 (16)
Therefore l₂ =8 x 2 =16 cm =0.16m
Magnetic force exerted on the wire,
F₂= BIl2
= 1.5 x 7 x 0.16 = 1.68 N
Hence, a force of 1.68 N acts in a vertically downward direction on the wire
See lessThe wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Current in both wires, I = 300 A Distance between the wires, r = 1.5 cm = 0.015 m Length of the two wires, l = 70 cm = 0.7 m Force between the two wires is given by the relation, F = μ0 I²/2πr Where, μ0= Permeability of free space = 4π x 10⁻⁷ Tm A⁻1 F = 4π x 10⁻⁷ x (300 )² / (2 π x 0.015) = 1.2 N/mRead more
Current in both wires, I = 300 A
Distance between the wires, r = 1.5 cm = 0.015 m
Length of the two wires, l = 70 cm = 0.7 m
Force between the two wires is given by the relation,
F = μ0 I²/2πr
Where,
μ0= Permeability of free space = 4π x 10⁻⁷ Tm A⁻1
F = 4π x 10⁻⁷ x (300 )² / (2 π x 0.015) = 1.2 N/m
Since the direction of the current in the wires is opposite, a repulsive force exists between them.
See lessA straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. (a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? (b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 m s⁻².
Length of the rod, 1 = 0.45 m Mass suspended by the wires, m = 60 g = 60 x 10-3 kg Acceleration due to gravity, g = 9.8 m/s2, Current in the rod flowing through the wire, I = 5A Ans (a). Magnetic field (B) is equal and opposite to the weight of the wire i.e., BIl=mg :.B =mg/Il = (60 x 10-3 x 9.8)/Read more
Length of the rod, 1 = 0.45 m
Mass suspended by the wires, m = 60 g = 60 x 10-3 kg
Acceleration due to gravity, g = 9.8 m/s2, Current in the rod flowing through the wire, I = 5A
Ans (a).
Magnetic field (B) is equal and opposite to the weight of the wire i.e.,
BIl=mg
:.B =mg/Il = (60 x 10-3 x 9.8)/ (5 x 0.45) =0.26 T
A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up in order to get zero tension in the wire. The magnetic field should be such that Fleming’s left hand rule gives an upward magnetic force.
Ans (b).
If the direction of the current is revered, then the force due to magnetic field and the weight of the wire acts in a vertically downward direction.
Total tension in the wire = Bll + mg
= 0.26 x 5 x 0.45 + (60 x 10-3) x 9.8 = 1.176 N
See lessA 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.
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