A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is _____ A-m$$^2$$. [Given $$\pi = 3.14$$]

We have a wire of length $$L = 314$$ cm $$= 3.14$$ m carrying a current of $$I = 14$$ A, bent to form a circle. We need to find the magnetic moment of this circular coil.

When the wire is bent into a circle, the circumference equals the length of the wire: $$2\pi r = L$$, so $$r = \dfrac{L}{2\pi} = \dfrac{3.14}{2 \times 3.14} = \dfrac{1}{2} = 0.5$$ m.

The area of the circular loop is $$A = \pi r^2 = 3.14 \times (0.5)^2 = 3.14 \times 0.25 = 0.785$$ m$$^2$$.

The magnetic moment of a current loop is $$M = nIA$$, where $$n$$ is the number of turns. Since the wire forms a single loop, $$n = 1$$. Therefore, $$M = 1 \times 14 \times 0.785 = 10.99 \approx 11$$ A-m$$^2$$.

Hence, the correct answer is 11.