91Ó°ÊÓ

The magnetic field at the center of a circular loop carrying a current I is given by the formula: \[B_A = \frac{\mu_0I}{2R}\] where: - \(B_A\) is the magnetic field at the center of the circle - \(\mu_0\) is the permeability of free space (\(4\pi × 10^{-7} Tm/A\)) - I is the current through the wire - R is the radius of the circle Since Wire A is bent into a circle of radius R, the circumference of the circle is equal to the length of Wire A: \(2 \pi R = \ell'\). We will use this to express R in terms of \(\ell'\), and then substitute it into the formula for \(B_A\).

From the given data: \[2 \pi R = \ell' \Rightarrow R = \frac{\ell'}{2\pi}\] Now substitute this value of R into the formula for \(B_A\): \[B_A = \frac{\mu_0 I}{2 \cdot \frac{\ell'}{2\pi}} = \frac{\pi\mu_0 I}{\ell'}\]

To find the magnetic field at the center of the square, we need to find the magnetic field produced by each side of the square at its center and then add the contributions from all four sides. The magnetic field produced by a straight current-carrying wire at a perpendicular distance \(r\) from it is given by: \[B = \frac{\mu_0 I}{2\pi r}\] For Wire B, each side has length \(a'\) and the total length of Wire B is also \(\ell'\), therefore: \[4 a' = \ell' \Rightarrow a' = \frac{\ell'}{4}\] The distance from the center of the square to the midpoint of each side is half of the diagonal, which is \(\frac{a'\sqrt{2}}{2}\). Therefore, the magnetic field produced by a single side of Wire B at its center is: \[B_{B_{side}} = \frac{\mu_0 I}{2\pi \cdot \frac{a'\sqrt{2}}{2}} = \frac{2\mu_0 I}{\pi a' \sqrt{2}}\] Since there are four sides in the square, the total magnetic field at the center of the square is: \[B_B = 4 \cdot B_{B_{side}} = 8 \cdot \frac{\mu_0 I}{\pi a' \sqrt{2}}\] Substitute the value of \(a'\) in terms of \(\ell'\): \[B_B = 8 \cdot \frac{\mu_0 I}{\pi \cdot \frac{\ell'}{4} \sqrt{2}} = \frac{8\pi\mu_0 I}{\ell' \sqrt{2}}\]

Now that we have expressions for both \(B_A\) and \(B_B\), we can find the ratio \(\frac{B_A}{B_B}\): \[\frac{B_A}{B_B} = \frac{\frac{\pi\mu_0 I}{\ell'}}{\frac{8\pi\mu_0 I}{\ell'\sqrt{2}}} = \frac{\pi\mu_0 I}{\ell'} \cdot \frac{\ell'\sqrt{2}}{8\pi\mu_0 I}\] Simplify the expression: \[\frac{B_A}{B_B} = \frac{\pi\mu_0 I\ell'\sqrt{2}}{8\pi\mu_0 I\ell'} = \frac{\pi^2}{8}\] The ratio \(\frac{B_A}{B_B}\) is equal to \(\frac{\pi^2}{8}\). Therefore, the correct answer is (D) \(\frac{\pi^2}{8}\).