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Step-by-Step Solution
Step 1: Understand the problem
We have two circular loops. The smaller loop (radius R₁ = 0.20 m) carries a current I = 2.0 A. The bigger loop (radius R₂ = 0.3 cm = 0.003 m) is placed coaxially at a distance d = 0.15 m from the smaller loop’s center. We want to find the magnetic flux (ϕ) linked with the bigger loop due to the magnetic field produced by the smaller loop.
Step 2: Write the formula for the magnetic field on the axis of a circular current loop
The magnetic field at a distance x from the center of a circular loop of radius R, carrying current I, along its axis is given by:
$B = \frac{\mu_0 \, I \, R^2}{2 \,\bigl(R^2 + x^2\bigr)^{\tfrac{3}{2}}}$
Here, μ₀ is the permeability of free space ($4\pi \times 10^{-7}\,\text{H\,m}^{-1}$). In our case, x = d = 0.15 m, and R = R₁ = 0.20 m.
Step 3: Calculate the magnetic field at the center of the bigger loop
Substitute I = 2.0 A, R = 0.20 m, x = 0.15 m into the formula for B:
$B = \frac{\mu_0 \,\times\, 2.0 \,\times\, (0.20)^2}{2 \,\bigl[(0.20)^2 + (0.15)^2\bigr]^{\tfrac{3}{2}}}
$
Step 4: Find the flux through the bigger loop
The magnetic flux ϕ through the bigger loop of radius R₂ is given by:
$ \phi = B \times \text{Area of bigger loop} = B \times \pi \,(R_2)^2
$
where R₂ = 0.003 m.
Step 5: Substitute numerical values and simplify
Putting all numerical factors together (and noting that μ₀ = $4\pi \times 10^{-7}\,\text{H\,m}^{-1}$), you obtain:
$
\phi
=
\frac{(4\pi \times 10^{-7}) \,\times\, 2.0 \,\times\, (0.20)^2}{2\,\bigl[(0.20)^2 + (0.15)^2\bigr]^{\tfrac{3}{2}}}
\,\times\,
\pi \,(0.003)^2
$
Simplifying carefully leads to a value close to:
$ \phi \approx 9.1 \times 10^{-11}\,\text{Wb.}
Step 6: State the final answer
Hence, the magnetic flux linked with the bigger loop is:
$\boxed{9.1 \times 10^{-11}\,\text{weber}}$
