Respuesta :
Answer:
a) The direction of the Induced current is clockwise and point outside the paper
b) Magnitude of the average induced emf = 107.5 mV
c) Average induced current = 43 mA
Explanation:
a) According to Lenz's law "the direction of the current induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes the initial changing magnetic field."
The direction of the Induced current is clockwise and point outside the paper
b) Magnitude of the average induced emf
B = 0.75 T
The magnitude of the induced emf is given by:
ε=-N(ΔΦ/Δt)
Initial diameter, d₁ = 20 cm = 0.2 m
Initial Radius , r₁ = 0.2/2 = 0.1 m
Initial Area, A₁ = πr₁² = π * (0.1)² = 0.01π m²
Final diameter, d₂ = 6 cm = 0.06 m
Final Radius , r₂ = 0.06/2 = 0.03 m
Final Area = A₂ = πr₂² = π * (0.03)² = 0.0009π m²
Since Φ=BA
Φ₁=BA₁ = 0.75 * 0.01π = 0.0236 Wb
Φ₂=BA₂ = 0.75 * 0.0009π = 0.0021 Wb
ε = -N(ΔΦ/Δt)
ΔΦ = Φ₂ - Φ₁ = 0.0021 - 0.0236 = -0.0215 Wb
Δt = 20.0cm = 0.2 m
Since it is a loop, N = 1
ε = -(-0.0215)/0.2
ε = 0.1075 V
ε = 107.5 mV
c) Average induced current
ε = IR
R = 2.5 Ω
I = ε /R
I = 0.1075/2.5
I = 0.043 A
I = 43 mA
(a) The direction of induced current will point out of the paper.
(b) The magnitude of the induced emf in the coil is 0.0429 V.
(c) The average induced current in the coil is 0.017 A.
The given parameters;
- magnetic field strength, B = 0.75 T
- initial diameter of the loop, d₁ = 20 cm
- final diameter of the loop, d₂ = 6 cm
- time of change, t = 0.5 s
The direction of the current can be determine by applying Flemings left hand rule, as the field points into the paper, the current will point out of the paper.
The magnitude of the induced emf is calculated by applying Faradays law;
[tex]emf = -N\frac{\Delta \phi}{dt} \\\\emf = N(\frac{\phi _1 - \phi_2}{t} )\\\\emf = N (\frac{BA_1\ - \ BA_2}{t} )\\\\emf = \frac{NB}{t} (A_1 - A_2)\\\\emf = \frac{NB}{t} (\frac{\pi d_1^2}{4} - \frac{\pi d_2^2}{4} )\\\\emf = \frac{NB\pi }{4t} (d_1^2 - d_2^2)\\\\emf = \frac{1 \times 0.75\times \pi}{4\times 0.5} (0.2^2 - 0.06^2)\\\\emf = 0.0429 \ V[/tex]
The average induced current in the coil is calculated as follows;
[tex]I = \frac{emf}{R} \\\\I = \frac{0.0429}{2.5} \\\\I = 0.017 \ A[/tex]
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