Step 1: Calculate the circumference (length) of the wire loop.
The circular loop has diameter $ D_{loop} = 10\text{ cm} = 0.10\text{ m} $, so its radius is $ R_{loop} = 0.05\text{ m} $. The length of the wire forming the loop is: \[ l = 2\pi R_{loop} = 2\pi(0.05) = 0.1\pi\text{ m} \]
Step 2: Calculate the cross-sectional area of the wire.
The wire has diameter $ d = 2\text{ mm} = 2 \times 10^{-3}\text{ m} $, so its radius is $ r_w = 10^{-3}\text{ m} $. The cross-section area is: \[ A_w = \pi r_w^2 = \pi \times 10^{-6}\text{ m}^2 \]
Step 3: Find the resistance of the wire loop.
Using $ R = \rho\, l / A_w $ with $ \rho = 2\times10^{-8}\ \Omega\text{m} $: \[ R = \frac{2\times10^{-8} \times 0.1\pi}{\pi\times10^{-6}} = \frac{2\times10^{-9}}{10^{-6}} = 2\times10^{-3}\ \Omega \]
Step 4: Find the required induced EMF.
By Ohm's law, the EMF needed to drive current $ I = 11\text{ A} $ through the loop is: \[ \varepsilon = IR = 11 \times 2\times10^{-3} = 2.2\times10^{-2}\text{ V} \]
Step 5: Apply Faraday's law to find dB/dt.
Since the field $ \vec{B} $ is perpendicular to the loop plane: \[ \varepsilon = A_{loop}\frac{dB}{dt}, \quad A_{loop} = \pi R_{loop}^2 = \pi(0.05)^2 = 2.5\pi\times10^{-3}\text{ m}^2 \] Therefore: \[ \frac{dB}{dt} = \frac{\varepsilon}{A_{loop}} = \frac{2.2\times10^{-2}}{2.5\pi\times10^{-3}} = \frac{22}{2.5\pi} \approx \frac{22}{7.854} \approx 2.8\text{ Ts}^{-1} \]
Step 6: State the final answer.
The rate at which the magnetic field must be changed is: \[ \boxed{2.8\text{ Ts}^{-1}} \]