Answer:
(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A
Explanation:
Given that,
The diameter of the loop, d = 40 cm
Radius, r = 20 cm
Initial magnetic field, B = 5 mT
Final magnetic field, B' = 55 mT
Initial magnetic flux,
[tex]\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb[/tex]
Final magnetic flux,
[tex]\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb[/tex]
Due to change in magnetic field an emf will be generated in the loop. It is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V[/tex]
Let I be the current in the loop. We can find it using Ohm's law such that,
[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A[/tex]
Hence, this is the required solution.
