Q73 / 75 · JEE Main 2026 (02 April Shift 1) · Physics
$1\,\mu$C charge moving with velocity $\vec{v} = \left(\hat{i} - 2\hat{j} + 3\hat{k}\right)$ m/s in the region of magnetic field $\vec{B} = \left(2\hat{i} + 3\hat{j} - 5\hat{k}\right)$ T. The magnitude of force acting on it is $\sqrt{\alpha} \times 10^{-6}$ N. The value of $\alpha$ is _______.
Correct answer: 171
Solution
The magnetic force acting on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Given:
$q = 1 \, \mu\text{C} = 10^{-6} \text{ C}$
$\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}$
$\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}$
Calculating the cross product $\vec{v} \times \vec{B}$:
$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$
$\vec{v} \times \vec{B} = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$
$\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}$
The magnitude of $\vec{v} \times \vec{B}$ is:
$|\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$
The magnitude of the force is:
$|\vec{F}| = q |\vec{v} \times \vec{B}| = 10^{-6} \times \sqrt{171} = \sqrt{171} \times 10^{-6} \text{ N}$
Comparing this with the given expression $\sqrt{\alpha} \times 10^{-6} \text{ N}$, we get:
$\alpha = 171$
Answer: $171$
Given:
$q = 1 \, \mu\text{C} = 10^{-6} \text{ C}$
$\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}$
$\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}$
Calculating the cross product $\vec{v} \times \vec{B}$:
$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$
$\vec{v} \times \vec{B} = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$
$\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}$
The magnitude of $\vec{v} \times \vec{B}$ is:
$|\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$
The magnitude of the force is:
$|\vec{F}| = q |\vec{v} \times \vec{B}| = 10^{-6} \times \sqrt{171} = \sqrt{171} \times 10^{-6} \text{ N}$
Comparing this with the given expression $\sqrt{\alpha} \times 10^{-6} \text{ N}$, we get:
$\alpha = 171$
Answer: $171$
Hence, the correct answer is 171.