Two concentric circular loops of radii $$r_1 = 30 \text{ cm}$$ and $$r_2 = 50 \text{ cm}$$ are placed in $$X-Y$$ plane as shown in the figure. A current $$I = 7 \text{ A}$$ is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately

Two concentric circular loops of radii $$r_1 = 30 \text{ cm}$$ and $$r_2 = 50 \text{ cm}$$ in the X-Y plane carry current $$I = 7 \text{ A}$$. We need to find the net magnetic moment.

The magnetic moment of a circular loop is:

$$\vec{M} = I \cdot A \cdot \hat{n}$$

where $$A = \pi r^2$$ is the area and $$\hat{n}$$ is the unit normal determined by the direction of current (right-hand rule).

From the figure, the inner loop ($$r_1 = 0.3 \text{ m}$$) carries current in one direction and the outer loop ($$r_2 = 0.5 \text{ m}$$) carries current in the opposite direction.

For the inner loop (current gives $$+\hat{k}$$):

$$M_1 = 7 \times \pi \times (0.3)^2 = 7 \times 0.09\pi = 0.63\pi \text{ A m}^2$$

For the outer loop (current gives $$-\hat{k}$$):

$$M_2 = 7 \times \pi \times (0.5)^2 = 7 \times 0.25\pi = 1.75\pi \text{ A m}^2$$

$$\vec{M}_{net} = 0.63\pi \hat{k} - 1.75\pi \hat{k} = -1.12\pi \hat{k}$$

$$\vec{M}_{net} = -1.12 \times 3.14 \hat{k} \approx -3.52 \hat{k} \approx -\dfrac{7}{2} \hat{k} \text{ A m}^2$$

The correct answer is Option B: $$-\dfrac{7}{2} \hat{k} \text{ A m}^2$$.