A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of the coil is parallel to the field. What is the emf of the coil

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

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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

answer:

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