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’.
A 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