A triangular shaped wire carrying 10 A current is placed in a uniform magnetic field of 0.5 T, as shown in figure. The magnetic force on segment CD is (Given BC = CD = BD = 5 cm).

We need to find the magnetic force $$F$$ acting on the segment $$CD$$ of the triangular wire loop placed in a uniform magnetic field.

The magnetic force on a straight current-carrying conductor of length $$L$$ carrying a current $$I$$ in a uniform magnetic field $$B$$ is given by the formula: $$F = I L B \sin\theta$$, where $$\theta$$ is the angle between the direction of the current element and the magnetic field vector.

From the given diagram, the triangular section $$\Delta BCD$$ is an equilateral triangle with sides $$BC = CD = BD = 5\text{ cm} = 0.05\text{ m}$$. Therefore, each internal angle of the triangle is $$60^\circ$$.

The magnetic field lines are directed horizontally from left to right, parallel to the segments $$AB$$ and $$DE$$. Therefore, the side $$BD$$ lies perfectly parallel to the magnetic field.

In the equilateral triangle, the side $$CD$$ slants downward towards the right. The angle between the direction of the current along segment $$CD$$ and the horizontal magnetic field lines is equal to the interior angle of the triangle at vertex $$D$$, which is $$\theta = 60^\circ$$.

We are given the following values:

Current, $$I = 10\text{ A}$$

Length of segment $$CD$$, $$L = 5\text{ cm} = 0.05\text{ m}$$

Magnetic field strength, $$B = 0.5\text{ T}$$

Angle, $$\theta = 60^\circ$$

Substituting these values into the magnetic force formula yields: $$F = 10 \times 0.05 \times 0.5 \times \sin(60^\circ)$$.

Since $$\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$$, the calculation becomes: $$F = 0.25 \times 0.866 = 0.2165\text{ N}$$.

Rounding to three decimal places, the magnitude of the force is approximately $$0.216\text{ N}$$.

Therefore, the correct answer is Option C: 0.216 N.