Answer:
0.24T/s
Explanation:
This problem can be solved if use the Lenz's Law, that is
[tex]\epsilon=-\frac{d\Phi_{B}}{dt}=-\frac{d(B*S)}{dt}[/tex] ( 1 )
but this emf is for the ring:
[tex]\epsilon=IR\\\epsilon=(2.0A)(1.5*10^{-3})=3.75*10^{-3}V[/tex]
and we can replace this value in ( 1 )
[tex]3.75*10^{-3}V=-S\frac{dB}{dt}=-\pi (0.07m)^{2}\frac{dB}{dt}\\\frac{dB}{dt}=\frac{3.75*10^{-3}V}{\pi (0.07m)^{2}}=0.24\frac{T}{s}[/tex]
where we have taken the area of a circle and a radius of 70cm=0.07m. dB/dt is the ratio of change of the magnetic field. Hence the answer is 0.24T/s
I hope this is useful for you
regards
