Answer:
[tex]I_{l} =44.84 \mu A[/tex]
Explanation:
given,
side of square loop = a = 2.10 cm
Resistance of the wire = 1.30×10⁻² Ω
Length of the loop = c = 1.10 cm
rate of increasing current = 130 A/s
[tex]\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))[/tex]
[tex]\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))[/tex]
[tex]I_{l} = \dfrac{V}{R}[/tex]
[tex]I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}[/tex]
[tex]I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))[/tex]
[tex]I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))[/tex]
[tex]I_{l} =44.84 \times 10^{-6}\A[/tex]
[tex]I_{l} =44.84 \mu A[/tex]
