Respuesta :
emf generated by the coil is 1.57 V
Explanation:
Given details-
Number of turns of wire- 1000 turns
The diameter of the wire coil- 1 cm
Magnetic field (Initial)= 0.10 T
Magnetic Field (Final)=0.30 T
Time=10 ms
The orientation of the axis of the coil= parallel to the field.
We know that EMF of the coil is mathematically represented as –
E=N(ΔФ/Δt)
Where E= emf generated
ΔФ= change inmagnetic flux
Δt= change in time
N= no of turns*area of the coil
Substituting the values of the above variables
=1000*3.14*0.5*10-4
=.0785
E=0.0785(.2/10*10-3)
=1.57 V
Thus, the emf generated is 1.57 V
The EMF of the coil of wire is; E = 1.57 V
We are given;
Number of turns of wire; N = 1000 turns
Diameter of the wire; d = 1 cm = 0.01 m
Radius; r = 0.01/2 = 0.005 m
Initial magnitude of magnetic field; B₁ = 0.10 T
Final magnitude of magnetic field; B₂ = 0.30 T
Time; Δt = 10 ms = 10 × 10⁻³ s
We are told that the axis of the coil is parallel to the field. This implies that the EMF will be maximum and θ = 0
According to Faraday's law of induction, the maximum induced EMF of the coil is expressed as;
E = N(Δ∅/Δt)
Where;
E is induced EMF
Δ∅ is change in magnetic flux
Δt is change in time
N is number of turns of coil
Now, Δ∅ = (B₂ - B₁)Acos θ
Since θ = 0°, then cos θ = 1. Thus;
Δ∅ = (B₂ - B₁)A
A is area = πr² = π(0.005²)
Thus;
E = (1000 × (0.3 - 0.1) × π(0.005²))/(10 × 10⁻³)
E = 1.57 V
Read more at; https://brainly.com/question/17113633
