Length of a side of the square coil, 1 = 10 cm = 0.1 m Current flowing in the coil, I = 12 A Number of turns on the coil, n = 20 Angle made by the plane of the coil with magnetic field, 0 = 30° Strength of magnetic field, B = 0.80 T Magnitude of the magnetic torque experienced by the coil in the magRead more
Length of a side of the square coil, 1 = 10 cm = 0.1 m
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic field, 0 = 30°
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
τ = n BIA sin0
Where, A = Area of the square coil = 1×1 = 0.1 x 0.1 = 0.01m²
So,
τ = 20 x 0.8 x 12 x 0.01 x Sin30°
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.
Length of the solenoid, 1 = 80 cm = 0.8 m There are five layers of windings of 400 turns each on the solenoid. Therefore the total number of turns on the solenoid, N = 5 x 400 = 2000 Diameter of the solenoid, D = 1.8 cm = 0.018 m Current carried by the solenoid, 1 = 8.0 A Magnitude of the magnetic fRead more
Length of the solenoid, 1 = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
Therefore the total number of turns on the solenoid, N = 5 x 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, 1 = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
B= μ0NI/l
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻1
B = (4π x 10⁻⁷ x 2000 x 8 )/0.8 = 2.5 x 10⁻² T
Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 x 10-2
Current flowing in wire A, Ia = 8.0 A Current flowing in wire B, Ib = 5.0 A Distance between the two wires, r = 4.0 cm = 0.04 m Length of a section of wire A, L = 10 cm = 0.1 m Force exerted on length L due to the magnetic field is given as: F = μ0 Ia Ib L/2 πr Where , μ0 = Permeability of free spacRead more
Current flowing in wire A, Ia = 8.0 A
Current flowing in wire B, Ib = 5.0 A
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, L = 10 cm = 0.1 m
Force exerted on length L due to the magnetic field is given as:
F = μ0 Ia Ib L/2 πr
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
F = (4π x 10⁻⁷ x 8 x 5 x 0.10 )/2 π x 0.04 = 2 x 10⁻⁵
The magnitude of force is 2 x 10-5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.
Length of the wire, 1 = 3 cm = 0.03 m Current flowing in the wire, I = 10 A Magnetic field, B = 0.27 T Angle between the current and magnetic field, 0 = 90° Magnetic force exerted on the wire is given as: F = BI/sin 0 = 0.27 x 10 x 0.03 sin90° = 8.1 x 10⁻2 N Hence, the magnetic force on the wire isRead more
Length of the wire, 1 = 3 cm = 0.03 m Current flowing in the wire, I = 10 A Magnetic field, B = 0.27 T Angle between the current and magnetic field, 0 = 90° Magnetic force exerted on the wire is given as: F = BI/sin 0 = 0.27 x 10 x 0.03 sin90° = 8.1 x 10⁻2 N Hence, the magnetic force on the wire is 8.1 x 10⁻2 N. The direction of the force can be obtained from Fleming’s left hand rule.
Current in the wire, I = 8 A Magnitude of the uniform magnetic field, B = 0.15 T Angle between the wire and magnetic field, 0 = 30°. Magnetic force per unit length on the wire is given as: f = BI sin0 = 0.15 x 8 x 1 x Sin30° = 0.6 N m1 Hence, the magnetic force per unit length on the wire is 0.6 N mRead more
Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, 0 = 30°. Magnetic force per unit length on the wire is given as: f = BI sin0 = 0.15 x 8 x 1 x Sin30° = 0.6 N m1
Hence, the magnetic force per unit length on the wire is 0.6 N m⁻1.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Length of a side of the square coil, 1 = 10 cm = 0.1 m Current flowing in the coil, I = 12 A Number of turns on the coil, n = 20 Angle made by the plane of the coil with magnetic field, 0 = 30° Strength of magnetic field, B = 0.80 T Magnitude of the magnetic torque experienced by the coil in the magRead more
Length of a side of the square coil, 1 = 10 cm = 0.1 m
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic field, 0 = 30°
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
τ = n BIA sin0
Where, A = Area of the square coil = 1×1 = 0.1 x 0.1 = 0.01m²
So,
τ = 20 x 0.8 x 12 x 0.01 x Sin30°
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.
See lessA closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Length of the solenoid, 1 = 80 cm = 0.8 m There are five layers of windings of 400 turns each on the solenoid. Therefore the total number of turns on the solenoid, N = 5 x 400 = 2000 Diameter of the solenoid, D = 1.8 cm = 0.018 m Current carried by the solenoid, 1 = 8.0 A Magnitude of the magnetic fRead more
Length of the solenoid, 1 = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
Therefore the total number of turns on the solenoid, N = 5 x 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, 1 = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
B= μ0NI/l
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻1
B = (4π x 10⁻⁷ x 2000 x 8 )/0.8 = 2.5 x 10⁻² T
Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 x 10-2
T.
See lessTwo long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Current flowing in wire A, Ia = 8.0 A Current flowing in wire B, Ib = 5.0 A Distance between the two wires, r = 4.0 cm = 0.04 m Length of a section of wire A, L = 10 cm = 0.1 m Force exerted on length L due to the magnetic field is given as: F = μ0 Ia Ib L/2 πr Where , μ0 = Permeability of free spacRead more
Current flowing in wire A, Ia = 8.0 A
Current flowing in wire B, Ib = 5.0 A
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, L = 10 cm = 0.1 m
Force exerted on length L due to the magnetic field is given as:
F = μ0 Ia Ib L/2 πr
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
F = (4π x 10⁻⁷ x 8 x 5 x 0.10 )/2 π x 0.04 = 2 x 10⁻⁵
The magnitude of force is 2 x 10-5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.
See lessA 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Length of the wire, 1 = 3 cm = 0.03 m Current flowing in the wire, I = 10 A Magnetic field, B = 0.27 T Angle between the current and magnetic field, 0 = 90° Magnetic force exerted on the wire is given as: F = BI/sin 0 = 0.27 x 10 x 0.03 sin90° = 8.1 x 10⁻2 N Hence, the magnetic force on the wire isRead more
Length of the wire, 1 = 3 cm = 0.03 m Current flowing in the wire, I = 10 A Magnetic field, B = 0.27 T
See lessAngle between the current and magnetic field, 0 = 90°
Magnetic force exerted on the wire is given as:
F = BI/sin 0 = 0.27 x 10 x 0.03 sin90° = 8.1 x 10⁻2 N
Hence, the magnetic force on the wire is 8.1 x 10⁻2 N. The direction of the force can be obtained from Fleming’s left hand rule.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?
Current in the wire, I = 8 A Magnitude of the uniform magnetic field, B = 0.15 T Angle between the wire and magnetic field, 0 = 30°. Magnetic force per unit length on the wire is given as: f = BI sin0 = 0.15 x 8 x 1 x Sin30° = 0.6 N m1 Hence, the magnetic force per unit length on the wire is 0.6 N mRead more
Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, 0 = 30°. Magnetic force per unit length on the wire is given as: f = BI sin0 = 0.15 x 8 x 1 x Sin30° = 0.6 N m1
Hence, the magnetic force per unit length on the wire is 0.6 N m⁻1.
See less