A current carrying circular loop of radius 2 cm with unit normal $$\hat{n} = \frac{\hat{k} + \hat{i}}{\sqrt{2}}$$ is placed in a magnetic field, $$\vec{B} = B_0(3\hat{i} + 2\hat{k})$$. If $$B_0 = 4 \times 10^{-3}$$ T and current $$I = 100\sqrt{2}$$ A, the torque experienced by the loop is __________ Wb.A. ($$\pi = 3.14$$)

Solution :

Radius of loop :

$$r = 2\text{ cm}$$

$$= 2 \times 10^{-2}\text{ m}$$

Area of loop :

$$A = \pi r^2$$

$$= 3.14 \times (2 \times 10^{-2})^2$$

$$= 12.56 \times 10^{-4}\text{ m}^2$$

Current in loop :

$$I = 100\sqrt2\text{ A}$$

Unit normal vector :

$$\hat n = \frac{\hat i + \hat k}{\sqrt2}$$

Magnetic field :

$$\vec B = B_0(3\hat i + 2\hat k)$$

Given,

$$B_0 = 4 \times 10^{-3}\text{ T}$$

Magnetic moment :

$$\vec m = IA\hat n$$

Torque on loop :

$$\tau = |\vec m \times \vec B|$$

$$= IA|\hat n \times \vec B|$$

Now,

$$\hat n \times \vec B=\frac{\hat i + \hat k}{\sqrt2}\times4 \times 10^{-3}(3\hat i + 2\hat k)$$

$$=\frac{4 \times 10^{-3}}{\sqrt2}[(\hat i + \hat k)\times(3\hat i + 2\hat k)]$$

Using,

$$\hat i \times \hat i = 0$$

$$\hat k \times \hat k = 0$$

$$\hat i \times \hat k = -\hat j$$

$$\hat k \times \hat i = \hat j$$

Therefore,

$$\hat n \times \vec B=\frac{4 \times 10^{-3}}{\sqrt2}(-2\hat j + 3\hat j)$$

$$=\frac{4 \times 10^{-3}}{\sqrt2}\hat j$$

Magnitude :

$$|\hat n \times \vec B|=\frac{4 \times 10^{-3}}{\sqrt2}$$

Hence,

$$\tau=(100\sqrt2)(12.56 \times 10^{-4})\left(\frac{4 \times 10^{-3}}{\sqrt2}\right)$$

$$=100 \times 12.56 \times 4 \times 10^{-7}$$

$$=5.024 \times 10^{-4}\text{ N m}$$

Final Answer :

$$5.024 \times 10^{-4}$$