(a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed ω in a magnetic field B directed perpendicular to the axis of rotation.
(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10–4 T and the angle of dip is 30° .
Answers
The potential difference developed between the ends of its wings is 1.25 V
Explanation:
An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10–4 T and the angle of dip is 30° .?
Solution:
Velocity of aeroplane = 9000 m/s
Span = 20 m
Horizontal component of the Earth’s magnetic field = 5 × 10–4 T
Angle of dip = 30°
e = Bv . V.I
e = 5 x 10^-4 x sin(30o) x 900 x 5 / 18 x 20
e = 1.25 V
Thus the potential difference developed between the ends of its wings is 1.25 V
Induced emf ϵ = ϵo sinωt
Potential difference = 1.44 V
Explanation:
(a) The working of an ac generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.
If N = number of turns in the coil
A = face area of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with filed B
ω = the angular frequency with which coil rotates
Magnetic flux linked with the coil at any instant t will be ∅ = NBA cosθ = NBA cos ωt
Induced emf = ϵ = -d∅/dt = -d(NBAcosωt) / dt = NBAω sin ωt
ϵ = ϵo sinωt (where ϵo = NBAω)
(b) Potential difference developed between the ends of the wings e = Bl
Given Velocity v = 900km/hour = 250m/s
The horizontal component of the Earth’s magnetic field = 5× 10-4T
Wing span (l) = 20 m
The vertical component of Earth’s magnetic field
BV = BH tan δ = 5×10⁻4 (tan 30° ) T
So the potential difference, e = 5*10⁻4 * (tan 30°) * 20 * 250 = 1.44 V