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

We begin by writing every given quantity in SI units.

The radius of the small loop is $$a = 0.3\ \text{cm}=0.3\times 10^{-2}\ \text{m}=3.0\times 10^{-3}\ \text{m}.$$

The radius of the large loop is $$R = 20\ \text{cm}=20\times 10^{-2}\ \text{m}=0.20\ \text{m}.$$

The distance between the two loop-centres measured along the common axis is $$d = 15\ \text{cm}=15\times 10^{-2}\ \text{m}=0.15\ \text{m}.$$

The current in the small loop is $$I = 2.0\ \text{A}.$$

Because the small loop is much smaller than every other length involved $$(a \ll d \ \text{and}\ a \ll R),$$ it is quite accurate to regard the small loop as a magnetic dipole situated on the axis. Its magnetic dipole moment is, by definition,

$$m = I\,\pi a^{2}.$$

A magnetic dipole produces an azimuthal vector potential $$A_{\phi}$$ at a point whose cylindrical co-ordinates relative to the dipole are $$(\rho ,\phi ,z)$$:

$$A_{\phi} = \frac{\mu_{0}}{4\pi}\, \frac{m\,\rho}{\left(\rho^{2}+z^{2}\right)^{3/2}}.$$

For the large loop we have $$\rho = R$$ and $$z = d$$, so its entire circumference lies on the circle where the above potential has the value

$$A_{\phi}(R,d) = \frac{\mu_{0}}{4\pi}\; \frac{m\,R}{\left(R^{2}+d^{2}\right)^{3/2}}.$$

The magnetic flux linked with the large loop equals the circulation of the vector potential around the loop (Stokes’ theorem):

$$\Phi = \oint \mathbf{A}\cdot d\mathbf{l} = 2\pi R\,A_{\phi}(R,d).$$

Substituting the expression for $$A_{\phi}$$ just written, we get

$$\Phi = 2\pi R \left( \frac{\mu_{0}}{4\pi}\; \frac{m\,R}{\left(R^{2}+d^{2}\right)^{3/2}} \right) = \frac{\mu_{0}\,m\,R^{2}}{2\left(R^{2}+d^{2}\right)^{3/2}}.$$

Now we replace $$m$$ by $$I\,\pi a^{2}$$:

$$\Phi = \frac{\mu_{0}\,I\,\pi a^{2}\,R^{2}} {2\left(R^{2}+d^{2}\right)^{3/2}}.$$

This expression contains only known quantities, so we substitute the numerical values one by one.

First, the constant $$\mu_{0}=4\pi \times 10^{-7}\ \text{H m}^{-1}.$$ Half of this is $$\frac{\mu_{0}}{2}=2\pi \times 10^{-7}\ \text{H m}^{-1}.$$

The factor $$\pi a^{2}$$ is

$$\pi a^{2}=\pi\,(3.0\times 10^{-3})^{2} =\pi\,(9.0\times 10^{-6}) =9\pi\times 10^{-6}\ \text{m}^{2}.$$

Multiplying by $$R^{2}$$ gives

$$\pi a^{2}R^{2}=\left(9\pi\times 10^{-6}\right)\left(0.20\right)^{2} =\left(9\pi\times 10^{-6}\right)\left(0.040\right) =0.36\pi\times 10^{-6}=3.6\pi\times 10^{-7}\ \text{m}^{4}.$$

Next we incorporate the current $$I=2\ \text{A}$$:

$$I\,\pi a^{2}R^{2}=2\times 3.6\pi\times 10^{-7} =7.2\pi\times 10^{-7}\ \text{A m}^{4}.$$

Multiplying by $$\mu_{0}/2$$ gives the whole numerator:

$$\left(\frac{\mu_{0}}{2}\right) \left(I\,\pi a^{2}R^{2}\right) =\left(2\pi\times 10^{-7}\right) \left(7.2\pi\times 10^{-7}\right) =14.4\pi^{2}\times 10^{-14}\ \text{Wb m}^{-1}.$$

Since $$\pi^{2}\approx 9.8696$$, the numerical value becomes

$$14.4\pi^{2}\times 10^{-14} =14.4\times 9.8696\times 10^{-14} \approx 142.1\times 10^{-14} =1.421\times 10^{-12}\ \text{Wb m}^{-1}.$$

We now turn to the denominator:

$$R^{2}+d^{2}=0.20^{2}+0.15^{2}=0.040+0.0225=0.0625,$$

and thus

$$\left(R^{2}+d^{2}\right)^{3/2} =(0.0625)^{3/2} =\sqrt{0.0625}\times 0.0625 =0.25\times 0.0625 =0.015625.$$

Finally, the flux is

$$\Phi =\frac{1.421\times 10^{-12}} {0.015625} \approx 9.09\times 10^{-11}\ \text{weber}.$$

Keeping only two significant figures,

$$\Phi\approx 9.1\times 10^{-11}\ \text{Wb}.$$

Comparing with the options supplied, this value corresponds to Option C.

Hence, the correct answer is Option C.