Question

A small, 2.00-mm-diameter circular loop with R = 1.00×10^−2 Ω is at the center of a large 100-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from +1.0A to −1.0A in 0.100 s . What is the induced current in the inner loop?

Answer 1

`i=2.1* 10^(-8)}\ A`

Step-by-step 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

`\varepsilon =-(d\phi )/(dt)`

`\varepsilon =-(BA )/(dt)`

`B=(\mu _oI)/(2\pi R)`

`\varepsilon =-\pi * r^2* (4\pi * 10^(-7)(I_2-I_1))/(2\pi Rdt)`

`\varepsilon =-\pi * 0.001^2* (4\pi * 10^(-7)(-1-1))/(2\pi * 0.05* 0.1)`

`\varepsilon =2.1* 10^(-10)\ V`

So induce current

i=emf/R

`i=(2.1* 10^(-10))/(10^(-2))\ A`

`i=2.1* 10^(-8)}\ A`