To solve this problem we must first find the potential on the body which is given as a product between the number of turns, the area and the variation of the magnetic field as a function of time. Once the potential is found, we will apply Ohm's Law with which we can find the induced current on the body. Our values are,
[tex]A = 7.00*10^{-4} m^2[/tex]
[tex]\Delta B = 3.30T-0.5T[/tex]
[tex]t = 0.99s[/tex]
[tex]R = 1.8\Omega[/tex]
a) The magnitude of average induced emf is given by
[tex]\epsilon_{emf} = NA (\frac{dB}{dt})[/tex]
Here N =1
[tex]\epsilon_{emf} = (1)(7.00 x 10^-4 m2)[\frac{(3.3 T - 0.5 T)}{(0.99s)}][/tex]
[tex]\epsilon_{emf}= 0.00196 V[/tex]
b) The magnitude of the induced current is
[tex]I = \frac{V}{R}[/tex]
Here Resistance is
[tex]R = 1.80\Omega[/tex]
[tex]I = \frac{(0.00196V)}{(1.80\Omega)}[/tex]
[tex]I = 0.00108 A[/tex]
Therefore the induced current is 0.00108A
