Question:medium

A $10 \, \text{cm$ long wire lies along $y$-axis. It carries a current of $1.0 \, \text{A}$ in positive $y$-direction. A magnetic field $\vec{B} = (5 \, \text{mT}) \, \hat{j} - (8 \, \text{mT}) \, \hat{k}$ exists in the region. The force on the wire is:}

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To calculate the magnetic force on a wire, use $\vec{F} = I (\vec{L} \times \vec{B})$. Expand the cross product carefully, considering the directions of the vectors involved.
Updated On: Jan 13, 2026
  • $(0.8 \, \text{mN}) \, \hat{i}$
  • $-(0.8 \, \text{mN}) \, \hat{i}$
  • $(80 \, \text{mN}) \, \hat{i}$
  • $-(80 \, \text{mN}) \, \hat{i}$
Show Solution

The Correct Option is B

Solution and Explanation

The magnetic force $\vec{F}$ on a current-carrying wire is defined as: \[ \vec{F} = I (\vec{L} \times \vec{B}), \] where the parameters are: \begin{itemize} \item $I = 1.0 \, \text{A}$ (current) \item $\vec{L} = (10 \, \text{cm}) \, \hat{j} = (0.1 \, \text{m}) \, \hat{j}$ (length vector of the wire) \item $\vec{B} = (5 \, \text{mT}) \, \hat{j} - (8 \, \text{mT}) \, \hat{k} = (5 \times 10^{-3}) \, \hat{j} - (8 \times 10^{-3}) \, \hat{k}$ (magnetic field) \end{itemize} Step 1: Compute the cross product $\vec{L} \times \vec{B}$:
\[ \vec{L} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0.1 & 0 \\ 0 & 5 \times 10^{-3} & -8 \times 10^{-3} \end{vmatrix}. \]
Expanding the determinant: \[ \vec{L} \times \vec{B} = \hat{i} \left( (0.1) \cdot (-8 \times 10^{-3}) - (0) \cdot (5 \times 10^{-3}) \right) - \hat{j} \left( (0) \cdot (-8 \times 10^{-3}) - (0) \cdot (0) \right) + \hat{k} \left( (0) \cdot (5 \times 10^{-3}) - (0.1) \cdot (0) \right). \]
Simplifying the expression: \[ \vec{L} \times \vec{B} = \hat{i} \left( -0.8 \times 10^{-3} \right) + 0 + 0. \] Therefore: \[ \vec{L} \times \vec{B} = -(0.8 \times 10^{-3}) \, \hat{i}. \]
Step 2: Calculate the force $\vec{F}$:
\[ \vec{F} = I (\vec{L} \times \vec{B}) = (1.0) \cdot (-(0.8 \times 10^{-3}) \, \hat{i}). \] Simplifying the result: \[ \vec{F} = -0.8 \, \text{mN} \, \hat{i}. \] The resulting force on the wire is: \[ \boxed{-(0.8 \, \text{mN}) \, \hat{i}}. \]
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