Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magnetic flux of $$10^{-3}$$ Wb passes through 'Q'. No current is passed through 'Q'. When no current passes through 'P' and a current of 2 A passes through 'Q', the flux through 'P' is:

According to the theory of mutual induction, when two coils are kept fixed with respect to each other, the magnetic flux in one coil produced by the current in the other is proportional to that current. The constant of proportionality is called the mutual inductance and is denoted by $$M$$. Mathematically we write the definition as

$$\Phi_{21}=M\,I_1 \qquad\text{and}\qquad \Phi_{12}=M\,I_2,$$

where

$$\Phi_{21}$$ is the flux linking coil 2 due to current $$I_1$$ in coil 1, and

$$\Phi_{12}$$ is the flux linking coil 1 due to current $$I_2$$ in coil 2.

In the present problem coil ‘P’ corresponds to coil 1, and coil ‘Q’ corresponds to coil 2. First, a current flows only in coil P and we observe the flux through coil Q.

The data given are

Current in coil P: $$I_P = 3\ \text{A},$$

Flux through coil Q caused by this current: $$\Phi_{Q}=10^{-3}\ \text{Wb}.$$

Using the defining formula $$\Phi_{Q}=M\,I_P$$, we can solve for the mutual inductance $$M$$:

$$M=\frac{\Phi_{Q}}{I_P}=\frac{10^{-3}\ \text{Wb}}{3\ \text{A}}.$$

Simplifying the ratio gives

$$M=\frac{1}{3}\times10^{-3}\ \text{H}=0.333\overline{3}\times10^{-3}\ \text{H}.$$

In scientific-notation form this is

$$M=3.33\overline{3}\times10^{-4}\ \text{H}.$$

Now we switch the situation: we set the current in coil P to zero and pass a current through coil Q. The numerical value of the mutual inductance does not change because it depends only on the geometry and relative placement of the two coils, not on which coil is carrying current.

The new current is

$$I_Q = 2\ \text{A},$$

and we want the resulting flux through coil P, which we call $$\Phi_{P}$$. Using the same formula but interchanging the coil labels, we write

$$\Phi_{P}=M\,I_Q.$$

Substituting the previously found $$M$$ and the given $$I_Q$$:

$$\Phi_{P}= \left(3.33\overline{3}\times10^{-4}\ \text{H}\right)\,(2\ \text{A}).$$

Carrying out the multiplication:

$$\Phi_{P}= 6.66\overline{6}\times10^{-4}\ \text{Wb}.$$

Rounding to three significant figures yields

$$\Phi_{P}=6.67\times10^{-4}\ \text{Wb}.$$

This numerical value exactly matches Option B.

Hence, the correct answer is Option B.