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

To solve for the force on a charge moving in a magnetic field, use the Lorentz force law: \(\vec{F} = q(\vec{v} \times \vec{B})\). Here, \(q = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C}\), \(\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}\) m/s, and \(\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}\) T. The cross product \(\vec{v} \times \vec{B}\) is calculated as follows: 

\(\vec{v} \times \vec{B} = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\1 & -2 & 3\\2 & 3 & -5\end{array}\right|\).

Use the determinant expansion: \(\vec{v} \times \vec{B} = \hat{i}((-2)(-5)-(3)(3)) - \hat{j}((1)(-5)-(3)(2)) + \hat{k}((1)(3)-(-2)(2))\).

This simplifies to:

\(\vec{v} \times \vec{B} = \hat{i}(10-9) - \hat{j}(-5-6) + \hat{k}(3+4)\).

\(\vec{v} \times \vec{B} = \hat{i}(1) - \hat{j}(-11) + \hat{k}(7)\).

\(\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}\).

Now, calculate the force: \(\vec{F} = q(\vec{v} \times \vec{B}) = 1 \times 10^{-6} (\hat{i} + 11\hat{j} + 7\hat{k}) \, \text{N}\).

The magnitude of the force is:

\(|\vec{F}| = 1 \times 10^{-6} \sqrt{1^2 + 11^2 + 7^2} \, \text{N}\).

\(|\vec{F}| = 1 \times 10^{-6} \sqrt{1 + 121 + 49} \, \text{N}\).

\(|\vec{F}| = 1 \times 10^{-6} \sqrt{171} \, \text{N}\).

The problem states the magnitude of the force as \(\sqrt{\alpha} \times 10^{-6} \, \text{N}\), so we have \(\sqrt{\alpha} = \sqrt{171}\) and thus \(\alpha = 171\).

The computed value of \(\alpha\) is 171, which satisfies the given range of 171,171.