Answer :
The coil of wire's EMF is E = 1.57.
We are given;
Wire's number of turns; N = 1000 turns
The wire's diameter is 1 cm (0.01 m).
Radius; r = 0.01/2 = 0.005 m
Initial magnetic field strength; B1 = 0.10 T , Final magnetic field strength; 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
What happens when a coil is in a magnetic field in a straight line?
The induced current is 0 when the rectangular coil is perpendicular to the magnetic field's direction. It's because the cross-section of the coil basically has zero or no magnetic field lines connected to it.
What is the induced emf when the coil's plane is parallel to the field?
The magnetic flux change is greatest when the coil plane is parallel to the magnetic field, and it is zero when the coil plane is normal to or perpendicular to the magnetic field, resulting in the maximum and smallest emf values, respectively.
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