A magnetic field induction is changing in magnitude at a constant...

The induced electromotive force in the circular loop is given by Faraday's law of induction:

$E = A \dfrac{dB}{dt} = \pi r^2 \dfrac{dB}{dt}$

The resistance of the copper wire forming the loop is:

$R = \rho \dfrac{l}{A'} = \rho \dfrac{2\pi r}{\pi a^2}$

The induced current in the loop is:

$I = \dfrac{E}{R} = \dfrac{\pi r^2 \dfrac{dB}{dt}}{\rho \dfrac{2\pi r}{\pi a^2}} = \dfrac{\pi r a^2}{2\rho} \dfrac{dB}{dt}$

The mass of the copper wire is the product of its volume and density:

$m = (\pi a^2)(2\pi r)\delta = 2\pi^2 a^2 r \delta$

Rearranging the above expression yields:

$\pi r a^2 = \dfrac{m}{2\pi \delta}$

Substituting this into the expression for the induced current, we get:

$I = \dfrac{1}{2\rho} \left( \dfrac{m}{2\pi \delta} \right) \dfrac{dB}{dt} = \dfrac{m}{4\pi \rho \delta} \dfrac{dB}{dt}$

Answer: $\dfrac{m}{4\pi\rho\delta}\dfrac{dB}{dt}$