Answer:
[tex]i=2.1\times 10^{-8}}\ A[/tex]
Explanation:
Given that
Diameter of small loop d= 2 mm or r=1 mm
Diameter of large loop D= 100 mm or R=50 mm
We know that induce emf given as
[tex]\varepsilon =-\dfrac{d\phi }{dt}[/tex]
[tex]\varepsilon =-\dfrac{BA }{dt}[/tex]
[tex]B=\dfrac{\mu _oI}{2\pi R}[/tex]
[tex]\varepsilon =-\pi \times r^2\times \dfrac{4\pi \times 10^{-7}(I_2-I_1)}{2\pi Rdt}[/tex]
[tex]\varepsilon =-\pi \times 0.001^2\times \dfrac{4\pi \times 10^{-7}(-1-1)}{2\pi \times 0.05\times 0.1}[/tex]
[tex]\varepsilon =2.1\times 10^{-10}\ V[/tex]
So induce current
i=emf/R
[tex]i=\dfrac{2.1\times 10^{-10}}{10^{-2}}\ A[/tex]
[tex]i=2.1\times 10^{-8}}\ A[/tex]
