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

Answer:

$i=2.1\times 10^{-8}}\ A$

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

$\varepsilon =-\dfrac{d\phi }{dt}$

$\varepsilon =-\dfrac{BA }{dt}$

$B=\dfrac{\mu _oI}{2\pi R}$

$\varepsilon =-\pi \times r^2\times \dfrac{4\pi \times 10^{-7}(I_2-I_1)}{2\pi Rdt}$

$\varepsilon =-\pi \times 0.001^2\times \dfrac{4\pi \times 10^{-7}(-1-1)}{2\pi \times 0.05\times 0.1}$

$\varepsilon =2.1\times 10^{-10}\ V$

So induce current

i=emf/R

$i=\dfrac{2.1\times 10^{-10}}{10^{-2}}\ A$

$i=2.1\times 10^{-8}}\ A$