Sides of the rectangular loop are 8 cm and 2 cm. Hence, area of the rectangular wire loop, A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2 Initial value of the magnetic field, B' = 0.3 T Rate of decrease of the magnetic field, dB/dt = 0.02 T/s Emf developed in the loop is given as: e = dφ/dt WherRead more
Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2
Initial value of the magnetic field, B’ = 0.3 T
Rate of decrease of the magnetic field, dB/dt = 0.02 T/s
Emf developed in the loop is given as: e = dφ/dt
Where, dφ= Change in flux through the loop area = AB
Therefore, e = d(AB)/dt = AdB/dt
= 16 x 10⁻⁴ x 0.02 = 0.32 x 10⁻⁴ V
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as: i = e/R
= (0.32 x 10⁻⁴)/1.6 = 2 x 10⁻⁵ A
Power dissipated in the loop in the form of heat is given as:
P = i2R
= (2 x 10⁻5)2 x 1.6 = 6.4 x 10⁻¹⁰ W The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.
Speed of the jet plane, v = 1800 km/h = 500 m/s Wing span of jet plane, l = 25 m Earth’s magnetic field strength, B = 5.0 x 10-4 T Angle of dip, δ = 30° Vertical component of Earth's magnetic field, Bv = B sin δ = 5 x 10_4sin 30° = 2.5 x 10-4 T Voltage difference between the ends of the wing can beRead more
Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 x 10-4 T
Angle of dip, δ = 30°
Vertical component of Earth’s magnetic field,
Bv = B sin δ = 5 x 10_4sin 30°
= 2.5 x 10-4 T
Voltage difference between the ends of the wing can be calculated as: e = (Bv) x l x v
= 2.5 x 10⁻⁴x 25 x 500 = 3.125 V
Hence, the voltage difference developed between the ends of the wings is 3.125 V.
Mutual inductance of a pair of coils, μ = 1.5 H Initial current, I₁= 0 A Final current I₂ = 20 A Change in current, dl = I2 — I₁= 20 — 0 = 20 A Time taken for the change, t = 0.5 s Induced emf e = dφ/dt ------------------Eq-1 Where, dφ is the change in the flux linkage with the coil. Emf is relatedRead more
Mutual inductance of a pair of coils, μ = 1.5 H
Initial current, I₁= 0 A Final current I₂ = 20 A
Change in current, dl = I2 — I₁= 20 — 0 = 20 A
Time taken for the change, t = 0.5 s
Induced emf e = dφ/dt ——————Eq-1
Where, dφ is the change in the flux linkage with the coil.
Initial current, I₁= 5.0 A Final current, I₂ = 0.0 A Change in current, dl = I₁ — I2 = 5 A Time taken for the change, t = 0.1 s Average emf, e = 200 V For self-inductance (L) of the coil, we have the relation for average emf as: e =di/dt L = e/(di/dt) =200/(5/0.1) = 4 H Hence, the self-induction ofRead more
Initial current, I₁= 5.0 A
Final current, I₂ = 0.0 A
Change in current, dl = I₁ — I2 = 5 A
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as: e =di/dt
Length of the wire, 1 = 10 m Falling speed of the wire, v = 5.0 m/s Magnetic field strength, B = 0.3 x 10-4 Wb m-2 Ans (a). Emf induced in the wire, e = Blv = 0.3 x 10-4 x 5 x 10 = 1.5 x 10-3 V Ans (b). Using Fleming's right hand rule, it can be inferred that the direction of the induced emf is fromRead more
Length of the wire, 1 = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 0.3 x 10-4 Wb m–2
Ans (a).
Emf induced in the wire, e = Blv = 0.3 x 10-4 x 5 x 10 = 1.5 x 10-3 V
Ans (b).
Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf is from West to East.
Ans (c).
The eastern end of the wire is at a higher potential.
Max induced emf = 0.603 V Average induced emf = 0 V Max current in the coil = 0.0603 A Average power loss = 0.018 W (Power comes from the external rotor] Radius of the circular coil, r = 8 cm = 0.08 m Area of the coil, A =πr2 = π x (0.08)2 m2 Number of turns on the coil, N = 20 Angular speed, ω= 50Read more
Max induced emf = 0.603 V
Average induced emf = 0 V
Max current in the coil = 0.0603 A
Average power loss = 0.018 W (Power comes from the external rotor]
Radius of the circular coil, r = 8 cm = 0.08 m
Area of the coil, A =πr2 = π x (0.08)2 m2
Number of turns on the coil, N = 20
Angular speed, ω= 50 rad/s
Magnetic field strength, B = 3 x 10⁻²T
Resistance of the loop, R = 10 Ω
Maximum induced emf is given as: e = Nω AB = 20 x 50 x-π x (0.08)2 x 3 x 10⁻²
= 0.603 V
The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero
Maximum current is given as:
I =e/R =0.603/10 = 0.0603 A
Average power loss due to joule heating:
P = eI/2 = (0.603×0.0603)/2= 0.018 W
The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.
Length of the rod, l = 1 m Angular frequency, ω = 400 rad/s Magnetic field strength, B = 0.5 T One end of the rod has zero linear velocity, while the other end has a linear velocity of lω Average linear velocity of the rod, V = (lω + 0 )/2 = lω/2 Emf developed between the centre and the ring, e = BlRead more
Length of the rod, l = 1 m
Angular frequency, ω = 400 rad/s
Magnetic field strength, B = 0.5 T
One end of the rod has zero linear velocity, while the other end has a linear velocity of lω
Average linear velocity of the rod, V = (lω + 0 )/2 = lω/2
Emf developed between the centre and the ring,
e = Blv = Bl (lω/2) = Bl²ω/2
= (0.5 x (l)²x400 )/2 =100V
Hence, the emf developed between the centre and the ring is 100 V.
Length of the rectangular wire, 1 = 8 cm = 0.08 m Width of the rectangular wire, b = 2 cm = 0.02 m Hence, area of the rectangular loop, A = lb = 0.08 x 0.02 = 16 x 10⁻4 m2 Magnetic field strength, B = 0.3 T Velocity of the loop, v = 1 cm/s = 0.01 m/s Ans (a). Emf developed in the loop is given as :Read more
Length of the rectangular wire, 1 = 8 cm = 0.08 m
Width of the rectangular wire, b = 2 cm = 0.02 m
Hence, area of the rectangular loop, A = lb = 0.08 x 0.02 = 16 x 10⁻4 m2
Magnetic field strength, B = 0.3 T
Velocity of the loop, v = 1 cm/s = 0.01 m/s
Ans (a).
Emf developed in the loop is given as :
e =Blv = 0.3 x 0.08 x 0.01
= 2.4 x 10⁻⁴ V
Time taken to travel along the width ,
l = Distance travelled /(Velocity) = b/v= 0.02 /0.01 = 2 s
Hence, the induced voltage is 2.4 x 10-4 V which lasts for 2 s.
Ans (b).
Emf developed, e = Bbv = 0.3 x 0.02 x 0.01 = 0.6 x 10⁻4 V
Time taken to travel along the length, t =Distance traveled /Velocity =l/v
=0.08/0.01 = 8 s
Hence, the induced voltage is 0.6 x 10-4 V which lasts for 8 s.
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2 Current carried by the solenoid changes from 2 A to 4 A. Therefore, change in current in the solenoid, di = 4- 2 = 2A Change iRead more
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid, di = 4- 2 = 2A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as: e =dφ/dt ————Eq-1
Where, φ= Induced flux through the small loop =BA————–Eq -2 B = Magnetic field =μ0ni————–Eq -3
μ0 = Permeability of free space = 4π x10-7 H/m
Hence, equation (i) reduces to:
e =d/dt (BA)
= Aμ0n (di/dt)
= 2 x 10⁻4 x 4π x10-7 x 1500 x 2/ 0.1
= 7.54 x 10⁻⁶ V
Hence, the induced voltage in the loop is 7.54 x 10-6V
Ans (a). As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic flux linked with it increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the antiRead more
Ans (a).
As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic
flux linked with it increases. According to Lenz’s law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the anti-clock wise direction. It means the direction of current will be along adcba.
Ans (b).
As the circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic field linked with it also decreases. By Lenz’s law, the induced current should produce a flux in the direction of original flux. For this, the induced current should flow in the anti-clock wise direction. It means the direction of current will be along a’d’c’b’.
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s⁻¹. If the cut is joined and the loop has a resistance of 1.6Ω how much power is dissipated by the loop as heat? What is the source of this power?
Sides of the rectangular loop are 8 cm and 2 cm. Hence, area of the rectangular wire loop, A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2 Initial value of the magnetic field, B' = 0.3 T Rate of decrease of the magnetic field, dB/dt = 0.02 T/s Emf developed in the loop is given as: e = dφ/dt WherRead more
Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2
Initial value of the magnetic field, B’ = 0.3 T
Rate of decrease of the magnetic field, dB/dt = 0.02 T/s
Emf developed in the loop is given as: e = dφ/dt
Where, dφ= Change in flux through the loop area = AB
Therefore, e = d(AB)/dt = AdB/dt
= 16 x 10⁻⁴ x 0.02 = 0.32 x 10⁻⁴ V
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as: i = e/R
= (0.32 x 10⁻⁴)/1.6 = 2 x 10⁻⁵ A
Power dissipated in the loop in the form of heat is given as:
P = i2R
= (2 x 10⁻5)2 x 1.6 = 6.4 x 10⁻¹⁰ W
See lessThe source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 x 10⁻⁴ T and the dip angle is 30°.
Speed of the jet plane, v = 1800 km/h = 500 m/s Wing span of jet plane, l = 25 m Earth’s magnetic field strength, B = 5.0 x 10-4 T Angle of dip, δ = 30° Vertical component of Earth's magnetic field, Bv = B sin δ = 5 x 10_4sin 30° = 2.5 x 10-4 T Voltage difference between the ends of the wing can beRead more
Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 x 10-4 T
Angle of dip, δ = 30°
Vertical component of Earth’s magnetic field,
Bv = B sin δ = 5 x 10_4sin 30°
= 2.5 x 10-4 T
Voltage difference between the ends of the wing can be calculated as: e = (Bv) x l x v
= 2.5 x 10⁻⁴x 25 x 500 = 3.125 V
Hence, the voltage difference developed between the ends of the wings is 3.125 V.
See lessA pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Mutual inductance of a pair of coils, μ = 1.5 H Initial current, I₁= 0 A Final current I₂ = 20 A Change in current, dl = I2 — I₁= 20 — 0 = 20 A Time taken for the change, t = 0.5 s Induced emf e = dφ/dt ------------------Eq-1 Where, dφ is the change in the flux linkage with the coil. Emf is relatedRead more
Mutual inductance of a pair of coils, μ = 1.5 H
Initial current, I₁= 0 A Final current I₂ = 20 A
Change in current, dl = I2 — I₁= 20 — 0 = 20 A
Time taken for the change, t = 0.5 s
Induced emf e = dφ/dt ——————Eq-1
Where, dφ is the change in the flux linkage with the coil.
Emf is related with mutual inductance as:
e = μ dI /dt —————————-Eq-2
Equating equations (1) and (2), we get
dφ/dt = μ dI /dt
dφ = 1.5x(20)
= 30 Wb
Hence, the change in the flux linkage is 30 Wb.
See lessCurrent in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit
Initial current, I₁= 5.0 A Final current, I₂ = 0.0 A Change in current, dl = I₁ — I2 = 5 A Time taken for the change, t = 0.1 s Average emf, e = 200 V For self-inductance (L) of the coil, we have the relation for average emf as: e =di/dt L = e/(di/dt) =200/(5/0.1) = 4 H Hence, the self-induction ofRead more
Initial current, I₁= 5.0 A
Final current, I₂ = 0.0 A
Change in current, dl = I₁ — I2 = 5 A
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as: e =di/dt
L = e/(di/dt)
=200/(5/0.1) = 4 H
Hence, the self-induction of the coil is 4 H.
See lessA horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s⁻¹ at right angles to the horizontal component of the earth’s magnetic field, 0.30 x 10⁻⁴ Wb m⁻². (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?
Length of the wire, 1 = 10 m Falling speed of the wire, v = 5.0 m/s Magnetic field strength, B = 0.3 x 10-4 Wb m-2 Ans (a). Emf induced in the wire, e = Blv = 0.3 x 10-4 x 5 x 10 = 1.5 x 10-3 V Ans (b). Using Fleming's right hand rule, it can be inferred that the direction of the induced emf is fromRead more
Length of the wire, 1 = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 0.3 x 10-4 Wb m–2
Ans (a).
Emf induced in the wire, e = Blv = 0.3 x 10-4 x 5 x 10 = 1.5 x 10-3 V
Ans (b).
Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf is from West to East.
Ans (c).
The eastern end of the wire is at a higher potential.
See lessA circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s⁻¹ in a uniform horizontal magnetic field of magnitude 3.0xl0⁻² T. Obtain the maximum and average emfinduced in the coil. If the coil forms a closed loop of resistance 10Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Max induced emf = 0.603 V Average induced emf = 0 V Max current in the coil = 0.0603 A Average power loss = 0.018 W (Power comes from the external rotor] Radius of the circular coil, r = 8 cm = 0.08 m Area of the coil, A =πr2 = π x (0.08)2 m2 Number of turns on the coil, N = 20 Angular speed, ω= 50Read more
Max induced emf = 0.603 V
Average induced emf = 0 V
Max current in the coil = 0.0603 A
Average power loss = 0.018 W (Power comes from the external rotor]
Radius of the circular coil, r = 8 cm = 0.08 m
Area of the coil, A =πr2 = π x (0.08)2 m2
Number of turns on the coil, N = 20
Angular speed, ω= 50 rad/s
Magnetic field strength, B = 3 x 10⁻²T
Resistance of the loop, R = 10 Ω
Maximum induced emf is given as: e = Nω AB = 20 x 50 x-π x (0.08)2 x 3 x 10⁻²
= 0.603 V
The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero
Maximum current is given as:
I =e/R =0.603/10 = 0.0603 A
Average power loss due to joule heating:
P = eI/2 = (0.603×0.0603)/2= 0.018 W
The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.
See lessA 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Length of the rod, l = 1 m Angular frequency, ω = 400 rad/s Magnetic field strength, B = 0.5 T One end of the rod has zero linear velocity, while the other end has a linear velocity of lω Average linear velocity of the rod, V = (lω + 0 )/2 = lω/2 Emf developed between the centre and the ring, e = BlRead more
Length of the rod, l = 1 m
Angular frequency, ω = 400 rad/s
Magnetic field strength, B = 0.5 T
One end of the rod has zero linear velocity, while the other end has a linear velocity of lω
Average linear velocity of the rod, V = (lω + 0 )/2 = lω/2
Emf developed between the centre and the ring,
e = Blv = Bl (lω/2) = Bl²ω/2
= (0.5 x (l)² x400 )/2 =100V
Hence, the emf developed between the centre and the ring is 100 V.
See lessA rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s⁻¹ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Length of the rectangular wire, 1 = 8 cm = 0.08 m Width of the rectangular wire, b = 2 cm = 0.02 m Hence, area of the rectangular loop, A = lb = 0.08 x 0.02 = 16 x 10⁻4 m2 Magnetic field strength, B = 0.3 T Velocity of the loop, v = 1 cm/s = 0.01 m/s Ans (a). Emf developed in the loop is given as :Read more
Length of the rectangular wire, 1 = 8 cm = 0.08 m
Width of the rectangular wire, b = 2 cm = 0.02 m
Hence, area of the rectangular loop, A = lb = 0.08 x 0.02 = 16 x 10⁻4 m2
Magnetic field strength, B = 0.3 T
Velocity of the loop, v = 1 cm/s = 0.01 m/s
Ans (a).
Emf developed in the loop is given as :
e =Blv = 0.3 x 0.08 x 0.01
= 2.4 x 10⁻⁴ V
Time taken to travel along the width ,
l = Distance travelled /(Velocity) = b/v= 0.02 /0.01 = 2 s
Hence, the induced voltage is 2.4 x 10-4 V which lasts for 2 s.
Ans (b).
Emf developed, e = Bbv = 0.3 x 0.02 x 0.01 = 0.6 x 10⁻4 V
Time taken to travel along the length, t =Distance traveled /Velocity =l/v
=0.08/0.01 = 8 s
Hence, the induced voltage is 0.6 x 10-4 V which lasts for 8 s.
(a] Emf developed in the loop is given as:
e=Blv = 0.3×0.08×0.01
= 2.4 x 10-4 V
See lessA long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2 Current carried by the solenoid changes from 2 A to 4 A. Therefore, change in current in the solenoid, di = 4- 2 = 2A Change iRead more
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 x 10⁻4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid, di = 4- 2 = 2A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as: e =dφ/dt ————Eq-1
Where, φ= Induced flux through the small loop =BA————–Eq -2 B = Magnetic field =μ0ni————–Eq -3
μ0 = Permeability of free space = 4π x10-7 H/m
Hence, equation (i) reduces to:
e =d/dt (BA)
= Aμ0n (di/dt)
= 2 x 10⁻4 x 4π x10-7 x 1500 x 2/ 0.1
= 7.54 x 10⁻⁶ V
Hence, the induced voltage in the loop is 7.54 x 10-6V
See lessUse Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.
Ans (a). As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic flux linked with it increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the antiRead more
Ans (a).
As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic
flux linked with it increases. According to Lenz’s law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the anti-clock wise direction. It means the direction of current will be along adcba.
Ans (b).
As the circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic field linked with it also decreases. By Lenz’s law, the induced current should produce a flux in the direction of original flux. For this, the induced current should flow in the anti-clock wise direction. It means the direction of current will be along a’d’c’b’.
See less