Radius of coil X, r₁= 16 cm = 0.16 m Radius of coil Y, r₂ = 10 cm = 0.1 m Number of turns of on coil X, n₁ =20 Number of turns of on coil Y, n₂ = 25 Current in coil X, I₁= 16 A Current in coil Y, I₂= 18 A Magnetic field due to coil X at their centre is given by the relation, B₁ = μ0 N₁l₁/2 r₁ Where,Read more
Radius of coil X, r₁= 16 cm = 0.16 m
Radius of coil Y, r₂ = 10 cm = 0.1 m
Number of turns of on coil X, n₁ =20
Number of turns of on coil Y, n₂ = 25
Current in coil X, I₁= 16 A
Current in coil Y, I₂= 18 A
Magnetic field due to coil X at their centre is given by the relation,
B₁ = μ0 N₁l₁/2 r₁
Where,μ0 = permeability of free space = 4 π x 10⁻⁴ T m A⁻¹
B₁ = (4 π x 10⁻⁴ x 25 x 18 )/( 2 x 0.10) = 9 π x 10⁻⁴ T (Towards West)
Hence ,net magnetic field can be obtained as , B = B₂ -B₁ = 9 π x 10⁻⁴ T – 4 π x 10⁻⁴ T =5 π x 10⁻⁴T
= 5 x 3.14 x 10⁻⁴ T = 1.57 x 10 ⁻³ T (Towards West)
Ans (a). Number of turns on the circular coil, n = 30 Radius of the coil, r = 8.0 cm = 0.08 m Area of the coil πr2 = π(0.08)2 = 0.0201 m2 Current flowing in the coil, I = 6.0 A Magnetic field strength, B = 1 T Angle between the field lines and normal with the coil surface, 0 = 60° The coil experiencRead more
Ans (a).
Number of turns on the circular coil, n = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil πr2 = π(0.08)2 = 0.0201 m2
Current flowing in the coil, I = 6.0 A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface, 0 = 60°
The coil experiences a torque in the magnetic field.
Hence, it turns.
The counter torque applied to prevent the coil from turning is given by the relation,
τ = n IBA Sin0 ———————Eq-1
= 30 x 6 x 1 x 0.0201 x Sin60º
= 3.133 Nm
Ans (b). It can be inferred from relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.
Magnetic field strength, B =6.5 x 10-4 T Charge of the electron, e = 1.6 x 10-19 C Mass of the electron, me = 9.1 x 10-31 kg Velocity of the electron, v = 4.8 x 106 m/s Radius of the orbit, r = 4.2 cm = 0.042 m Frequency of revolution of the electron = v Angular frequency of the electron = ω = 2πv VRead more
Magnetic field strength, B =6.5 x 10-4 T
Charge of the electron, e = 1.6 x 10-19 C
Mass of the electron, me = 9.1 x 10-31 kg
Velocity of the electron, v = 4.8 x 106 m/s
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = v
Angular frequency of the electron = ω = 2πv
Velocity of the electron is related to the angular frequency as: v = rω
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force.
Hence, we can write:
mv²/r = evB
=> eB = mv/r =m (rω)/r = m (r.2πv)/r
=> v = Be/2πm
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as
v = ( 6.5 x 10-4 x 1.6 x 10-19 )/( 2 x 3.14 x 9.1 x 10-31
= 1.82 x 10⁶ Hz or approx. 18MHz
Hence ,the frequency of the electron is around 18MHz and is independent of the speed of the electron.
Magnetic field strength, B = 6.5 G = 6.5 x 10⁻4 T Speed of the electron, v = 4.8 x 106 m/s Charge on the electron, e = 1.6 x 10-19 C Mass of the electron, me = 9.1 x 10 -31 kg Angle between the shot electron and magnetic field, 0 = 90° Magnetic force exerted on the electron in the magnetic field isRead more
Magnetic field strength, B = 6.5 G = 6.5 x 10⁻4 T
Speed of the electron, v = 4.8 x 106 m/s
Charge on the electron, e = 1.6 x 10-19 C
Mass of the electron, me = 9.1 x 10 –31 kg
Angle between the shot electron and magnetic field, 0 = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sin0
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
Fe = mv/r²
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
Fe = F
=> mv/r² = evB sin0
=> r = mv/ evB sin0
So,
r = (9.1 x 10 –31 x 4.8 x 106 )/( 6.5 x 10⁻4 x 1.6 x 10-19 x Sin 90º)
= 4.2 x 10⁻² m = 4.2 cm
Hence ,the radius of the circular orbit of the electron is 4.2 cm.
For moving coil meter M₁: Resistance, R₁ = 10 QΩ Number of turns, N₁ = 30 Area of cross-section, A₁ = 3.6 x 10⁻3 m2 Magnetic field strength, B₁ = 0.25 T Spring constant K₁ = K For moving coil meter M₂: Resistance, R₂= 14 Q Number of turns, N₂= 42 Area of cross-section, A₂ = 1.8 x 10⁻3 m2 Magnetic fiRead more
For moving coil meter M₁:
Resistance, R₁ = 10 QΩ
Number of turns, N₁ = 30
Area of cross-section, A₁ = 3.6 x 10⁻3 m2
Magnetic field strength, B₁ = 0.25 T
Spring constant K₁ = K
For moving coil meter M₂:
Resistance, R₂= 14 Q
Number of turns, N₂= 42
Area of cross-section, A₂ = 1.8 x 10⁻3 m2
Magnetic field strength, B₂ = 0.50 T
Spring constant, K₂ = K
Ans (a).
Current sensitivity of M₁ is given by
Is₁ = N₁B₁A₁/K₁
and, current sensitivity of M₂ is given by
Is₂ = N₂B₂A₂/K₂
Therefore Ratio Is₂ /Is₁ = N₂B₂A₂ /N₁B₁A₁ (Since Spring constant, K₂ = K)
=> Is₂ /Is₁ = (42 x 0.5 x 1.8 x 10⁻3 x K)/(K x 30 x 0.25 x 3.6 x 10⁻3 ) = 1.4
Hence ,the ratio of current sensitivity of M₂ to M₁ is 1.4
Ans (b).
Voltage sensitivity for M₂ is given by
Vs₂ = N₂B₂A₂/ K₂ R₂
And Volatage sensitivity for M₁ is given by
Vs₁ = N₁B₁A₁/K₁R₁
Therefore Ratio Vs₂/ Vs₁ = N₂B₂A₂ K₁R₁/ K₂ R₂ N₁B₁A₁
Vs₂/ Vs₁ = (42 x 0.5 x 1.8 x 10⁻³ x 10 x K)/(K x 14 x 30 x 0.25 x 3.6 x 10⁻³ )
=> Vs₂/ Vs₁ = 1
Hence ,the ratio of voltage sensitivity of M₂ to M₁ is 1.
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.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Radius of coil X, r₁= 16 cm = 0.16 m Radius of coil Y, r₂ = 10 cm = 0.1 m Number of turns of on coil X, n₁ =20 Number of turns of on coil Y, n₂ = 25 Current in coil X, I₁= 16 A Current in coil Y, I₂= 18 A Magnetic field due to coil X at their centre is given by the relation, B₁ = μ0 N₁l₁/2 r₁ Where,Read more
Radius of coil X, r₁= 16 cm = 0.16 m
Radius of coil Y, r₂ = 10 cm = 0.1 m
Number of turns of on coil X, n₁ =20
Number of turns of on coil Y, n₂ = 25
Current in coil X, I₁= 16 A
Current in coil Y, I₂= 18 A
Magnetic field due to coil X at their centre is given by the relation,
B₁ = μ0 N₁l₁/2 r₁
Where,μ0 = permeability of free space = 4 π x 10⁻⁴ T m A⁻¹
B₁ = (4 π x 10⁻⁴ x 25 x 18 )/( 2 x 0.10) = 9 π x 10⁻⁴ T (Towards West)
Hence ,net magnetic field can be obtained as ,
See lessB = B₂ -B₁ = 9 π x 10⁻⁴ T – 4 π x 10⁻⁴ T =5 π x 10⁻⁴T
= 5 x 3.14 x 10⁻⁴ T = 1.57 x 10 ⁻³ T (Towards West)
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
Ans (a). Number of turns on the circular coil, n = 30 Radius of the coil, r = 8.0 cm = 0.08 m Area of the coil πr2 = π(0.08)2 = 0.0201 m2 Current flowing in the coil, I = 6.0 A Magnetic field strength, B = 1 T Angle between the field lines and normal with the coil surface, 0 = 60° The coil experiencRead more
Ans (a).
Number of turns on the circular coil, n = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil πr2 = π(0.08)2 = 0.0201 m2
Current flowing in the coil, I = 6.0 A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface, 0 = 60°
The coil experiences a torque in the magnetic field.
Hence, it turns.
The counter torque applied to prevent the coil from turning is given by the relation,
τ = n IBA Sin0 ———————Eq-1
= 30 x 6 x 1 x 0.0201 x Sin60º
= 3.133 Nm
Ans (b).
See lessIt can be inferred from relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.
In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Magnetic field strength, B =6.5 x 10-4 T Charge of the electron, e = 1.6 x 10-19 C Mass of the electron, me = 9.1 x 10-31 kg Velocity of the electron, v = 4.8 x 106 m/s Radius of the orbit, r = 4.2 cm = 0.042 m Frequency of revolution of the electron = v Angular frequency of the electron = ω = 2πv VRead more
Magnetic field strength, B =6.5 x 10-4 T
Charge of the electron, e = 1.6 x 10-19 C
Mass of the electron, me = 9.1 x 10-31 kg
Velocity of the electron, v = 4.8 x 106 m/s
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = v
Angular frequency of the electron = ω = 2πv
Velocity of the electron is related to the angular frequency as: v = rω
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force.
Hence, we can write:
mv²/r = evB
=> eB = mv/r =m (rω)/r = m (r.2πv)/r
=> v = Be/2πm
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as
v = ( 6.5 x 10-4 x 1.6 x 10-19 )/( 2 x 3.14 x 9.1 x 10-31
= 1.82 x 10⁶ Hz or approx. 18MHz
Hence ,the frequency of the electron is around 18MHz and is independent of the speed of the electron.
See lessIn a chamber, a uniform magnetic field of 6.5 G (1 G = 10⁻⁴ T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s⁻¹ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10⁻¹⁹ C, me = 9.1×10⁻³¹ kg)
Magnetic field strength, B = 6.5 G = 6.5 x 10⁻4 T Speed of the electron, v = 4.8 x 106 m/s Charge on the electron, e = 1.6 x 10-19 C Mass of the electron, me = 9.1 x 10 -31 kg Angle between the shot electron and magnetic field, 0 = 90° Magnetic force exerted on the electron in the magnetic field isRead more
Magnetic field strength, B = 6.5 G = 6.5 x 10⁻4 T
Speed of the electron, v = 4.8 x 106 m/s
Charge on the electron, e = 1.6 x 10-19 C
Mass of the electron, me = 9.1 x 10 –31 kg
Angle between the shot electron and magnetic field, 0 = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sin0
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
Fe = mv/r²
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
Fe = F
=> mv/r² = evB sin0
=> r = mv/ evB sin0
So,
r = (9.1 x 10 –31 x 4.8 x 106 )/( 6.5 x 10⁻4 x 1.6 x 10-19 x Sin 90º)
= 4.2 x 10⁻² m = 4.2 cm
Hence ,the radius of the circular orbit of the electron is 4.2 cm.
See lessTwo moving coil meters, M₁ and M₂ have the following particulars: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³m², B₁ = 0.25 T R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m² , B₂ = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
For moving coil meter M₁: Resistance, R₁ = 10 QΩ Number of turns, N₁ = 30 Area of cross-section, A₁ = 3.6 x 10⁻3 m2 Magnetic field strength, B₁ = 0.25 T Spring constant K₁ = K For moving coil meter M₂: Resistance, R₂= 14 Q Number of turns, N₂= 42 Area of cross-section, A₂ = 1.8 x 10⁻3 m2 Magnetic fiRead more
For moving coil meter M₁:
Resistance, R₁ = 10 QΩ
Number of turns, N₁ = 30
Area of cross-section, A₁ = 3.6 x 10⁻3 m2
Magnetic field strength, B₁ = 0.25 T
Spring constant K₁ = K
For moving coil meter M₂:
Resistance, R₂= 14 Q
Number of turns, N₂= 42
Area of cross-section, A₂ = 1.8 x 10⁻3 m2
Magnetic field strength, B₂ = 0.50 T
Spring constant, K₂ = K
Ans (a).
Current sensitivity of M₁ is given by
Is₁ = N₁B₁A₁/K₁
and, current sensitivity of M₂ is given by
Is₂ = N₂B₂A₂/K₂
Therefore Ratio Is₂ /Is₁ = N₂B₂A₂ /N₁B₁A₁ (Since Spring constant, K₂ = K)
=> Is₂ /Is₁ = (42 x 0.5 x 1.8 x 10⁻3 x K)/(K x 30 x 0.25 x 3.6 x 10⁻3 ) = 1.4
Hence ,the ratio of current sensitivity of M₂ to M₁ is 1.4
Ans (b).
Voltage sensitivity for M₂ is given by
Vs₂ = N₂B₂A₂/ K₂ R₂
And Volatage sensitivity for M₁ is given by
Vs₁ = N₁B₁A₁/K₁R₁
Therefore Ratio Vs₂/ Vs₁ = N₂B₂A₂ K₁R₁/ K₂ R₂ N₁B₁A₁
Vs₂/ Vs₁ = (42 x 0.5 x 1.8 x 10⁻³ x 10 x K)/(K x 14 x 30 x 0.25 x 3.6 x 10⁻³ )
=> Vs₂/ Vs₁ = 1
Hence ,the ratio of voltage sensitivity of M₂ to M₁ is 1.
See lessA 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