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.
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 less