(a) 25 A
The induced emf in the circuit is given by
[tex]\epsilon = -\frac{\Delta \Phi}{\Delta t}[/tex]
where
[tex]\Delta Phi[/tex] is the variation of magnetic flux through the bracelet
[tex]\Delta t = 19.0 ms =0.019 s[/tex] is the time interval
The variation of magnetic flux is
[tex]\Delta \Phi = A \Delta B[/tex]
where
[tex]A=0.005 m^2[/tex] is the area enclosed by the bracelet
[tex]\Delta B=0.81 T-2.70 T=-1.89 T[/tex] is the change of magnetic field strength
So we find
[tex]\Delta \Phi = (0.005 m^2)(-1.89 T)=-9.45\cdot 10^{-3} Wb[/tex]
And so the induced emf is
[tex]\epsilon = -\frac{-9.45\cdot 10^{-3} Wb}{0.019 s}=0.50 V[/tex]
Since the resistance of the bracelet is
[tex]R=0.02\Omega[/tex]
the induced current is
[tex]I=\frac{V}{R}=\frac{0.50 V}{0.02 \Omega}=25 A[/tex]
(b) 12.5 W
The power delivered to the bracelet is given by
[tex]P=V I[/tex]
where we have
I = 25 A is the current
V = 0.50 V is the emf induced in the bracelet
Substituting numbers, we find
[tex]P=(0.50 V)(25 A)=12.5 W[/tex]
