Question

A small, 1.80-mm-diameter circular loop with R = 1.30×10^−2 Ω is at the center of a large 110-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:

current is 4.4723 × $10^{-8}$ A

Explanation:

given data

diameter = 1.80 mm

R = 1.30×10^−2 Ω

diameter = 110 mm

current in the outer loop = +1.0A to −1.0A

time = 0.100 sec

to find out

induced current

solution

we know here emf that is

emf = d∅ dt

emf = -d(BA) / dt

here B =  μi / 2πr

so emf = πr² / t ×  μ / 2r × (i1 - i2 )

put the value

emf = π(1.8/2×10^-3)² / 0.1 ×  4 π × $10^{-7}$ / 2(110/2×10^-3) × (1 - (-1) )

emf = 5.8140 × $10^{-10}$ V

so current is

current = emf / R

current = 5.8140 × $10^{-10}$ / 1.30×10^−2

current = 4.4723 × $10^{-8}$ A

Answer 2

The current induced in the inner loop is 4.47 x 10⁻⁸ A.

The given parameters;

• diameter of the larger loop, d = 110 mm
• radius of the larger loop, r = 55 mm = 0.055 m
• resistance of the circular loop, R = 1.3 x 10⁻² Ω
• time of current change, t = 0.1 s
• current in the loop, I₁ = +1 .0 A,  I₂ = -1.0 A
• diameter of the inner loop = 1.8 mm
• radius of the inner loop = 0.9 mm

The induced emf in the loop is calculated as follows;

$emf = \frac{dBA}{dt} \\\\emf = \frac{\mu_o \Delta i \times (\pi r^2)}{2 r \Delta t} \\\\emf = \frac{(4\pi \times 10^{-7} \times (1-(-1)) \times (\pi \times 0.0009^2)}{2 \times 0.055 \times 0.1} \\\\emf =5.82\times 10^{-10} \ V$

The current induced in the inner loop is calculated as follows;

$V = IR\\\\I = \frac{V}{R} = \frac{5.82\times 10^{-10}}{1.3 \times 10^{-2}} \\\\I = 4.47 \times 10^{-8} \ A$

Thus, the current induced in the inner loop is 4.47 x 10⁻⁸ A.

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