Answer:
the emf will be [tex]2.4(1.20\times 10^{-2}+13.4\times 10^{-5}t^3)[/tex]
Explanation:
We have given radius of the coil r = 3.6 cm = 0.036 m
So area [tex]A=\pi r^2=3.14\times 0.036^2=0.0040m^2[/tex]
Number of turns N = 600
Emf is given by [tex]e=N\frac{d\Phi }{dt}[/tex]
[tex]e=N\frac{d\Phi }{dt}=NA\frac{dB}{dt}=600\times 0.0040\times d\frac{1.20\times 10^{-2}t+3.35\times 10^{-5}t^4}{dt}=2.4(1.20\times 10^{-2}+13.4\times 10^{-5}t^3)[/tex]
So the emf will be [tex]2.4(1.20\times 10^{-2}+13.4\times 10^{-5}t^3)[/tex]