Question

A circular loop of radius 0.3cm lies parallel to a much bigger circular loop of radius 20cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15cm. If a current of 2.0A flows through the smaller loop, then the flux linked with bigger loop is

$\begin{align}
  & (A)6\times {{10}^{-11}}Wb \\
 & (B)3.3\times {{10}^{-11}}Wb \\
 & (C)6.6\times {{10}^{-9}}Wb \\
 & (D)9.1\times {{10}^{-11}}Wb \\
\end{align}$

Answer 1

Hint: For calculating the magnetic flux first calculate the total magnetic field. Because magnetic flux is defined as the number of magnetic field lines passing through a given closed surface. Thus magnetic flux is the product of the magnetic field and its area. then by substituting the value of B and A we will get the magnetic flux.
Formula used:
Flux formula, $\Phi =BA$
here, B is the magnetic field and A is the area.
B = $\dfrac{{{\mu }_{_{0}}}I{{R}^{2}}}{2{{(R+x)}^{\dfrac{3}{2}}}}$ and $A=\pi {{r}^{2}}$

Complete Answer:
Circular loop radius, r =0.3cm =$3\times {{10}^{-3}}m$
Bigger circular loop radius, R=20cm=0.2m
Distance, x = 15cm=0.15m
Current flowing, I=2.0A
Here we have to find the flux. For this first we have to calculate B,
B = $\dfrac{{{\mu }_{_{0}}}I{{R}^{2}}}{2{{(R+x)}^{\dfrac{3}{2}}}}$=$\dfrac{4\pi \times {{10}^{-7}}\times 2\times {{0.2}^{2}}}{2{{({{0.2}^{2}}+{{0.15}^{2}})}^{\dfrac{3}{2}}}}=3.22\times {{10}^{-6}}$T
Then,
$\Phi =BA$ = $3.22\times {{10}^{-6}}$ $\times 3.14\times {{(3\times {{10}^{-3}})}^{2}}$ =9.1$\times {{10}^{-11}}Wb$

Thus the option (D) is correct.

Additional information:
Magnetic flux is defined because the amount of magnetic flux the world perpendicular to the magnetic flux’s direction. It’s measurement is taken in Teslas.

Note:
The foremost significant difference between magnetic flux and therefore the magnetic flux is that the magnetic flux is that the region round the magnet where the moving charge experiences a force. Whereas the magnetic flux the strength of the magnetic lines produced by the magnet.