Question

The magnetic field perpendicular to a single 13.2-cm diameter circular loop of copper wire decreases uniformly from 0.670 T to zero. If the wire is 2.25 mm in diameter, how much charge moves past a point in the coil during this operation? (rhoCu = 1.68 x 10-8 Ω.m)

Answer 1

Answer:

5.23 C

Explanation:

The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.

Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁

B₁ = 0.670 T and B₂ = 0 T

ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T

A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m

A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²

ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt

Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire =  1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m

R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω

So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω  

IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C

Since ΔQ = It = 5.232 C ≅ 5.23 C

So the charge is 5.23 C

Answer 2

The induced emf through wire depends on the current flow (indirectly with charge flow as well).

The value of charge moving past a point in the coil during its operations is 5.23 C.

What is the magnetic field?

The region in a space where a particle experiences some magnetic force, is known as the magnetic field.

Given data -

The diameter of the circular loop is, d = 13.2 cm = 0.132 m.

The change in magnetic field strength is, ΔB = 0.670 T.

The new diameter of the wire is, d' = 2.25 mm = 2.25 × 10³ m.

The resistivity of the wire is, $\rho = 1.68 \times 10^{-8} \;\rm \Omega.m$.

The current in the wire is given by the following expression,

I = ε/R

Here,

ε is the induced emf in the wire.

R is the resistance of the wire.

And the expression for the induced emf of the wire is given as,

ε = -ΔΦ/Δt

Here,

ΔΦ is the change in magnetic flux. And its expression is,

ε = A × ΔB

Here,

A is the area of the loop. And its value is, A = πd²/4.

Solving as,

A = π(0.132 m)²/4

A = 0.01368 m²
A = `1.368 × 10⁻² m²

Now, calculating the induced emf as,

ε = ΔΦ/Δt

ε = A × ΔB/Δt

ε = 1.368 × 10⁻² m² × (0.670 T)/Δt

ε = 0.9166 × 10⁻² Tm²/Δt

Now, the resistance R of the circular wire is,

R = ρ × L/A'

Here,

L is the length of the wire and its value is. L = πd .

And A' is the new cross-sectional area of wire,

A' = πd'²/4

So, the resistance is,

R = ρ × L/A'

R = 1.68 x 10⁻⁸  ×( π × 0.132 m) ÷ π(2.25 × 10⁻³ m)²/4 =

R = 0.88704/5.0625

R = 0.1752 × 10⁻² Ω

R = 1.752 × 10⁻³ Ω

Now, the current flow (I) in the wire is given as,

I = ε/R

I = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω  

And obtaining the value of charge from the expression of current as,

Q = IΔt
Q = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω

Q = 0.5232 × 10 C

Q = 5.23 C

Thus, we can conclude that the value of charge moving past a point in the coil during its operations is 5.23 C.

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