Question

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm.The centre of the smaller loop is on the axis of the bigger loop.the distance between their centre is 15 cm.If a current of 2.0 A flows through the bigger loop,then the flux linked the smaller loop is.....
Answer is 9.1x10^-11 Wb
Plz solved out it in simple and easy method

Answer 1

your ans is in attachment  refer it

Answer 2

Answer:

The flux on the loop is $9.1 \times 10^{-11} \mathrm{Wb}$

Given data:

Circular loop radius, r = 0.3 cm = $3 \times 10^{-3} \ m$

Bigger circular loop radius, R = 20 cm = 0.2 m

Distance, x = 15 cm = 0.15 m

Current flowing, I = 2.0 A

To find:

Flux = ?

Solution:

Apply the flux formula given below:

$\mathrm{Flux}=\frac{\mu_{0} \mathrm{IR}^{2} \mathrm{r}^{2} \pi}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$

$=\frac{4 \pi \times 10^{-7} \times 2.0 \times 0.2^{2} \times\left(3 \times 10^{-3}\right)^{2} \times \pi}{2\left(0.2^{2}+0.15^{2}\right)^{3 / 2}}$

$=\frac{2.8447 \times 10^{-12}}{0.03125}$

$=91.03 \times 10^{-12}$

$\Rightarrow \text { Flux }=9.1 \times 10^{-11} \ \mathrm{Wb}$