Answer:
a) [tex]1.007 * 10^{-7}[/tex] V.m , into the page
b) [tex]-5.39 * 10^{-8}[/tex] V, counterclockwise
Explanation:
a) [tex]d\Phi = B.dA\\\\B_r = \frac{\mu_0I}{2\pi r}\\\\dA = b.dr\\\\\Phi = \int\limits^b_a {B.dA} = \int\limits^{a+d}_d {\frac{\mu_0I}{2\pi r}.b.dr} =\frac{\mu_0I}{2\pi }.b.ln(\frac{a+d}{d} )\\[/tex]
At t = 2.90, I = 3.62 + 1.49 * (2.90)^2 = 16.15 A
[tex]\Phi = \frac{4\pi * 10^{-7}*16.15*0.027}{2\pi} ln(1.46/0.46) = 1.007 * 10^{-7}[/tex] V.m
Direction is into the page.
b) [tex]emf = -d\Phi/dt = -\frac{\mu_0}{2\pi }.b.ln(\frac{a+d}{d}).(2.98t)= \\=-\frac{4\pi * 10^{-7}*0.027}{2\pi} ln(1.46/0.46)*2.98*2.90= -5.39 * 10^{-8} V[/tex]
Direction is counterclockwise.
