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 _____.
Correct Answer: 32
Solution and Explanation
To find the force \( \vec{F} \) acting on the charge, we use the equation for the magnetic force on a moving charge: \(\vec{F} = q(\vec{v} \times \vec{B})\). Here, \( q = 4.0 \, \mu \text{C} = 4.0 \times 10^{-6} \, \text{C} \), \( \vec{v} = 4.0 \times 10^6 \, \text{ms}^{-1} \hat{j} \), and \(\vec{B} = 2\hat{k} \, \text{T}\).
The cross product \(\vec{v} \times \vec{B}\) can be determined using the determinant method:
\(\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 4.0 \times 10^6 & 0 \\ 0 & 0 & 2 \end{vmatrix}\)
Calculating the determinant, we find:
\(\vec{v} \times \vec{B} = \hat{i}(4.0 \times 10^6 \times 2) - \hat{j}(0 \times 0) + \hat{k}(0 \times 0) = 8.0 \times 10^6 \hat{i} \, \text{m/s}\).
Substituting back, the force is:
\(\vec{F} = (4.0 \times 10^{-6}) (8.0 \times 10^6 \hat{i}) = 32 \hat{i} \, \text{N}\).
Thus, the value of \( x \) is 32, which confirms it fits within the given range [32,32].
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