Question

A \(1\,\mu\text{C}\) charge moving with velocity} \[ \vec{v} = (\hat{i} - 2\hat{j} + 3\hat{k}) \, \text{m/s} \] in the region of magnetic field} \[ \vec{B} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \, \text{T} \] The magnitude of force acting on it is \( \sqrt{\alpha} \times 10^{-6} \) N. The value of \(\alpha\) is ____.}

Answer 1

Correct Answer: 171

Concept: Magnetic force on a moving charge: \[ \vec{F} = q(\vec{v} \times \vec{B}) \] Magnitude: \[ F = q |\vec{v} \times \vec{B}| \]
Step 1:Compute cross product} \[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix} \] \[ = \hat{i}((-2)(-5) - 3 \cdot 3) - \hat{j}(1(-5) - 3 \cdot 2) + \hat{k}(1 \cdot 3 - (-2)\cdot2) \] \[ = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4) \] \[ = \hat{i} + 11\hat{j} + 7\hat{k} \]
Step 2:Find magnitude} \[ |\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} \] \[ = \sqrt{1 + 121 + 49} \] \[ = \sqrt{171} \]
Step 3:Calculate force \[ q = 1\mu C = 10^{-6} C \] \[ F = 10^{-6}\sqrt{171} \] Thus \[ F = \sqrt{171} \times 10^{-6} \text{ N} \] \[ \boxed{\alpha = 171} \]