Question 6.6: 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 3.0×10 −2 T. Obtain the maximum and average emf induced 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?
Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-230
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Given,
Radius of the circular coil,r = 8.0 cm = 0.08 m
No. of turns in the coil, N =20
Angular frequency ω = 50 rad/s
Magnitude of magnetic field B =3.0 × 10^-2 T
Resistance of the closed loop = 10 Ω
The area of the coil can be calculated using the formula
A =πr² = 3.14 × (0.08 m)²
The maximum induced e.m.f is calculated as follows:e = NωAB
e = 20 × 50 rad/s × 3.14 × (0.08 m)² × 3.0 × 10^–2 T
e = 0.603 V
For a complete cycle, the average induced emf will be zero.
The maximum current for the circular coil can be calculated as follows:
I = e/R
Substituting the values
I = 0.603 V/10 Ω
I = 0.0603A
The average power loss due to Joule’s heating effect in the circular coil will be given by:
P = (ei)/2
P = (0.603 V × 0.0603 A)/2
P = 0.018 W.
Radius of the circular coil,r = 8.0 cm = 0.08 m
No. of turns in the coil, N =20
Angular frequency ω = 50 rad/s
Magnitude of magnetic field B =3.0 × 10^-2 T
Resistance of the closed loop = 10 Ω
The area of the coil can be calculated using the formula
A =πr² = 3.14 × (0.08 m)²
The maximum induced e.m.f is calculated as follows:e = NωAB
e = 20 × 50 rad/s × 3.14 × (0.08 m)² × 3.0 × 10^–2 T
e = 0.603 V
For a complete cycle, the average induced emf will be zero.
The maximum current for the circular coil can be calculated as follows:
I = e/R
Substituting the values
I = 0.603 V/10 Ω
I = 0.0603A
The average power loss due to Joule’s heating effect in the circular coil will be given by:
P = (ei)/2
P = (0.603 V × 0.0603 A)/2
P = 0.018 W.
Answered by
2
Answer:
Given,
Radius of the circular coil,r = 8.0 cm = 0.08 m
No. of turns in the coil, N =20
Angular frequency ω = 50 rad/s
Magnitude of magnetic field B =3.0 × 10^-2 T
Resistance of the closed loop = 10 Ω
The area of the coil can be calculated using the formula
A =πr² = 3.14 × (0.08 m)²
The maximum induced e.m.f is calculated as follows:e = NωAB
e = 20 × 50 rad/s × 3.14 × (0.08 m)² × 3.0 × 10^–2 T
e = 0.603 V
For a complete cycle, the average induced emf will be zero.
The maximum current for the circular coil can be calculated as follows:
I = e/R
Substituting the values
I = 0.603 V/10 Ω
I = 0.0603A
The average power loss due to Joule’s heating effect in the circular coil will be given by:
P = (ei)/2
P = (0.603 V × 0.0603 A)/2
P = 0.018 W.
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