Electromagnetic Induction and Alternating Current

Magnetic force on asingle coil = BIL
0.01 X 5.0 X 10^-3 X 8 X  10^-3
4.0 X 10^-7
Torque acting on the coil = 20 x 4.0 x 10^-7 x 8.0 x 10^-3= 6.4 x 10^-8 N m

Area = 30 x 150 = 4500 m². Flux = ɸ= BA = 5 x 10-4 x 4500 = 2.25 Wb. So induced emf is E = Δɸ / Δt = 2.25 / 1= 2.25 V.

The magnitude of the induced emf ε = dΦB / dt = - d NBA / dt = - N B dA / dt = - 100 x 5 x 10^-2 x 0.2 / 0.05 = 20 volt.

R = 0.1 Ω; ø = 6 t²+ 7 t + 1 ; t = 1 s; e = ? I = ?
e = dø/dt = 6 x 2 t + 7 = 12 t + 7
∴At t = 1, e = 12 + 7 = 19 mV
Now I = e / R= 19 x 10^-3/ 0.1 = 19 x 10^-2 = 190 mA.

Number of turns of 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 × 10⁻⁴ m²_. Current carried by the solenoid changes from 2 A to 4 A. Change in current in the solenoid, di = 4 − 2 = 2 A. Change in time, dt = 0.1 s. Induced emf in the solenoid is given by Faraday’s law as:
e = dΦ / dt
where, Φ = BA (B = μ0ni)
μ0 = Permeability of free space = 4π×10⁻⁷ H/m
e = d BA/dt = A μ0 n di/dt = 2×10−4 × 4π×10⁻⁷ × 1500 × 2/0.1 = 7.54 ×10⁻⁶ V

L = 0.88 m; B1= 2.5 Wb/m²; dt = 2 s; B2= 3 Wb/m²; e = ?
ø1= A B1=πr²B1
; Here, 2πr = L = 0.88
∴r = 0.88 / 2π
∴ø1=πx 0.88 x 0.88 x 2.5 / 4 π²
= 0.22 x 0.88 x 2.5 / 3.14 = 0.1540
ø2= A B2= L²B2
; Here L = 0.88 / 4 = 0.22
∴ø2= 0.22 x 0.22 x 3 = 0.1452
∴dø = 0.1540–0.1452 = 0.0088
∴e = dø / dt = 0.0088 / 2 = 0.0044 V = 4.4 mV

Area of the loop, A = 4 cm x 4 cm = 16 x 10^{- 4} m² ,
Magnetic field, B = 5 mT = 5 x 10^{- 3} T,
Angle, θ = 00,
Magnetic flux through the loop, Φ = BA cos θ = 5 x 10^{- 3} x 16 x 10^{- 4} cos 0= 8 x 10^{- 6} Wb

Initial flux through the coil, Φ = BA cosθ = 3.0 × 10–5 × (π ×10–2) × cos 0º = 3π × 10–7 Wb. Final flux after the rotation, Φ= 3.0 × 10–5× (π×10–2) × cos 180° = –3π× 10–7 Wb. Therefore, estimated value of the induced emf is, ε= N ΔΦ / Δt= 500 × (6π× 10–7)/0.25 = 3.8 × 10–3VI = ε / R= 1.9 × 10–3 A

The potential difference between the ends of the axle of a car is given by ε = Bvvl = 3.0 x 10^{- 5} x 20 x 1.5 = 9.0 x 10^{- 4} V.

Δ (NΦ) = MΔ I1 = 1.5 x (20 – 0) = 30 Wb

Wing span l = 50 m. Speed v = 200 m/sHorizontal component of earth's field BH = 0.3 gauss = 0.3 x 10^{- 4} T. Vertical component of earth's field BV = 0.5 gauss = 0.5 x 10^{- 4} T. An aeroplane flying horizontally cuts only the vertical component of the earth's field. So, the potential difference between the wing tips of the aeroplane is ε = Bvvl = 0.5 x 10^{- 4} x 200 x 50 = 0.5 V

Peak emf is given byε₀ = NBAω
N = 100, B = 0.10 T, A = 10 cm² = 10 x 10^{- 4} m², ω = 100 rad/s
So, ε₀ = NBAω = 100 x 0.10 x 10 x 10^{- 4} x 100 = 1 V .
Peak current through the external resistance = ε₀ / (10 + 10) = 1/20 = 0.05 A.

The reading of the millivoltmeter = motional emf produced across the axle = BvlSpeed v = 135 km/h = 135 x 5/18 m/s = 37.5 m/sMagnetic field B = 0.2 G = 0.2 x 10^{- 4} T Length l = 1 m Therefore, the reading of the millivoltmeter = 0.2 x 10^{- 4} x 37.5 x 1 = 0.00075 V = 0.75 mV

The network of resistors forms a balanced Wheatstone bridge,  and hence resistor of 3 ohm in the middle can be neglected. Remaining resistors have an equivalent resistance = 3 ohm. Total resistance = 3 + 1 = 4 ohm. Induced emf, Bvl = IR Or 2 x v x 0.10 = 1 x 10^{- 3} x 4. This gives v = 2 cm/sec.

Length of rod l = 1 m,  Speed v = 72 km/h = 20 m/s, Angle of dip δ = 300. A rod falling vertically along east-west direction cuts only the horizontal component of the earth's field. Vertical component of earth's field Bv = 1/2 √3 gauss = (1/2 √3) x 10^{- 4} T. Horizontal component of earth's field BH = Bv / tan δ = (1/2 √3)/(1 / √3) gauss = 0.5 x 10^-4 T.  So, the potential difference between the ends of is ε = BHvl = 0.5 x 10^{- 4} x 20 x 1 = 10^{- 3} V