atrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & v_z \\ B_x & B_y & B_z \end{vmatrix}$$

$$\vec{F} = 1 \times 10^{-6} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$$

Expanding the determinant:

$$\vec{F} = 1 \times 10^{-6}(\hat{i}[(-2)(-5) - (3)(3)] - \hat{j}[(1)(-5) - (3)(2)] + \hat{k}[(1)(3) - (-2)(2)])$$

$$\vec{F} = 1 \times 10^{-6}[\hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)]$$

$$\vec{F} = 1 \times 10^{-6}[1\hat{i} + 11\hat{j} + 7\hat{k}]$$

So, the magnitude $|\vec{F}|$ is:

$$|\vec{F}| = 10^{-6} \sqrt{1^2 + 11^2 + 7^2}$$

$$|\vec{F}| = 10^{-6} \sqrt{1 + 121 + 49}$$

$$|\vec{F}| = \sqrt{171} \times 10^{-6}\ \mathrm{N}$$

$$\Rightarrow \sqrt{\alpha} \times 10^{-6}\ \mathrm{N} = \sqrt{171} \times 10^{-6}\ \mathrm{N}$$

$$\Rightarrow \alpha = 171$$

Therefore, the value of $\alpha$ is $171$. Thus, the correct answer is $171$.

19:T561,

चुंबकीय क्षेत्र $\vec{B}$ में $\vec{v}$ वेग से गतिमान आवेश $q$ पर लगने वाला चुंबकीय बल इस प्रकार दिया जाता है :

$$\vec{F} = q(\vec{v} \times \vec{B})$$

$$\vec{F} = q \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\\ v_x & v_y & v_z \\\\ B_x & B_y & B_z \end{vmatrix}$$

$$\vec{F} = 1 \times 10^{-6} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\\ 1 & -2 & 3 \\\\ 2 & 3 & -5 \end{vmatrix}$$

सारणिक का प्रसार करने पर:

$$\vec{F} = 1 \times 10^{-6}(\hat{i}[(-2)(-5) - (3)(3)] - \hat{j}[(1)(-5) - (3)(2)] + \hat{k}[(1)(3) - (-2)(2)])$$

$$\vec{F} = 1 \times 10^{-6}[\hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)]$$

$$\vec{F} = 1 \times 10^{-6}[1\hat{i} + 11\hat{j} + 7\hat{k}]$$

अतः परिमाण $|\vec{F}|$ है:

$$|\vec{F}| = 10^{-6} \sqrt{1^2 + 11^2 + 7^2}$$

$$|\vec{F}| = 10^{-6} \sqrt{1 + 121 + 49}$$

$$|\vec{F

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