Magnitude of Magnetic Force Acting on Moving Charge

To find the magnitude of the magnetic force acting on the charge, we use the Lorentz magnetic force formula: $\vec{F} = q \left(\vec{v} \times \vec{B}\right)$

Given:

• Charge, $q = 1\text{ }\mu\text{C} = 10^{-6}\text{ C}$
• Velocity, $\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}\text{ m/s}$
• Magnetic field, $\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}\text{ T}$

First, we calculate the cross product $\vec{v} \times \vec{B}$: $\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$

Expanding the determinant: $\vec{v} \times \vec{B} = \hat{i}\left[(-2)(-5) - (3)(3)\right] - \hat{j}\left[(1)(-5) - (3)(2)\right] + \hat{k}\left[(1)(3) - (-2)(2)\right]$ $\vec{v} \times \vec{B} = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$ $\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}$

Next, we calculate the magnitude of $\vec{v} \times \vec{B}$: $\left|\vec{v} \times \vec{B}\right| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$

Now, the magnitude of the magnetic force $F$ is: $F = q \left|\vec{v} \times \vec{B}\right| = 10^{-6} \times \sqrt{171}\text{ N} = \sqrt{171} \times 10^{-6}\text{ N}$

Comparing this with the given expression $F = \sqrt{\alpha} \times 10^{-6}\text{ N}$, we get: $\alpha = 171$