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.
Current 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 less