Question

A charge of \( 4.0 \, \mu \text{C} \) is moving with a velocity of \( 4.0 \times 10^6 \, \text{ms}^{-1} \) along the positive y-axis under a magnetic field \( \vec{B} \) of strength \( (2\hat{k}) \, \text{T} \). The force acting on the charge is \( x \hat{i} \, \text{N} \). The value of \( x \) is _____.

Answer 1

Correct Answer: 32

The magnetic force \( \vec{F} \) on a moving charge is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \). Given values are \( q = 4.0 \, \mu \text{C} = 4.0 \times 10^{-6} \, \text{C} \), \( \vec{v} = 4.0 \times 10^6 \, \text{ms}^{-1} \hat{j} \), and \(\vec{B} = 2\hat{k} \, \text{T}\).

The cross product \(\vec{v} \times \vec{B}\) is calculated using the determinant:

\(\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 4.0 \times 10^6 & 0 \\ 0 & 0 & 2 \end{vmatrix}\)

Evaluating the determinant yields:

\(\vec{v} \times \vec{B} = \hat{i}(4.0 \times 10^6 \times 2) - \hat{j}(0 \times 0) + \hat{k}(0 \times 0) = 8.0 \times 10^6 \hat{i} \, \text{m/s}\).

Substituting these values back into the force equation:

\(\vec{F} = (4.0 \times 10^{-6}) (8.0 \times 10^6 \hat{i}) = 32 \hat{i} \, \text{N}\).

Therefore, the value of \( x \) is 32, which falls within the specified range [32,32].