A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be $$\sqrt{x} \times 10^{-5}$$ Nm. The value of $$x$$ is _________

We have an equilateral-triangle coil carrying a current in a uniform magnetic field. For a current-carrying loop, the magnitude of the magnetic torque is given by the formula

$$\tau = N\,I\,A\,B\,\sin\phi,$$

where $$N$$ is the number of turns, $$I$$ is the current, $$A$$ is the area of the coil, $$B$$ is the magnetic-field magnitude, and $$\phi$$ is the angle between the field $$\vec B$$ and the normal to the plane of the coil.

Only one turn is mentioned, so we set $$N = 1.$$

The coil is an equilateral triangle of side $$a = 10\text{ cm} = 0.1\text{ m}.$$ The area of an equilateral triangle is

$$A = \frac{\sqrt{3}}{4}\,a^{2}.$$

Substituting $$a = 0.1\text{ m}$$, we get

$$A = \frac{\sqrt{3}}{4}\,(0.1)^{2} = \frac{\sqrt{3}}{4}\,(0.01) = 0.0025\,\sqrt{3}\;\text{m}^{2}.$$

The magnetic field is horizontal with magnitude $$B = 20\text{ mT} = 20\times10^{-3}\text{ T} = 2\times10^{-2}\text{ T}.$$

The current in the coil is $$I = 0.2\text{ A}.$$

The question states that the plane of the coil becomes parallel to the magnetic field. When the plane of the coil is parallel to $$\vec B$$, the normal to the plane is perpendicular to $$\vec B$$, giving $$\phi = 90^{\circ}$$ and hence $$\sin\phi = 1.$$

Now we substitute every quantity into the torque formula:

$$\tau = (1)\,(0.2)\,(0.0025\sqrt{3})\,(2\times10^{-2})\,(1).$$

First multiply $$0.2$$ and $$0.0025$$:

$$0.2 \times 0.0025 = 0.0005.$$

Next multiply this by $$2\times10^{-2}$$:

$$0.0005 \times 2\times10^{-2} = (0.0005 \times 2)\times10^{-2} = 0.001\times10^{-2} = 1\times10^{-5}.$$

Finally, attach the factor $$\sqrt{3}$$:

$$\tau = \sqrt{3} \times 10^{-5}\;\text{N\,m}.$$

The problem states the torque in the form $$\sqrt{x}\times10^{-5}\text{ N\,m}$$. Comparing coefficients, we have

$$x = 3.$$

So, the answer is $$3.$$