Question

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:}

• A. $(0.8 \, \text{mN}) \, \hat{i}$
• B. $-(0.8 \, \text{mN}) \, \hat{i}$
• C. $(80 \, \text{mN}) \, \hat{i}$
• D. $-(80 \, \text{mN}) \, \hat{i}$

Hint:

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.

Answer 1

The Correct Option is B

The magnetic force $\vec{F}$ on a current-carrying wire is given by: \[ \vec{F} = I (\vec{L} \times \vec{B}), \] where: \begin{itemize} \item $I = 1.0 \, \text{A}$ is the current, \item $\vec{L} = (10 \, \text{cm}) \, \hat{j} = (0.1 \, \text{m}) \, \hat{j}$ is the length vector of the wire, \item $\vec{B} = (5 \, \text{mT}) \, \hat{j} - (8 \, \text{mT}) \, \hat{k} = (5 \times 10^{-3}) \, \hat{j} - (8 \times 10^{-3}) \, \hat{k}$. \end{itemize} Step 1: Compute $\vec{L \times \vec{B}$:}
\[ \vec{L} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & 0.1 & 0
0 & 5 \times 10^{-3} & -8 \times 10^{-3} \end{vmatrix}. \]
Expand the determinant: \[ \vec{L} \times \vec{B} = \hat{i} \left( 0.1 \cdot (-8 \times 10^{-3}) - 0 \cdot (5 \times 10^{-3}) \right) - \hat{j} \left( 0 \cdot (-8 \times 10^{-3}) - 0 \cdot 0 \right) + \hat{k} \left( 0 \cdot (5 \times 10^{-3}) - 0.1 \cdot 0 \right). \]
Simplify: \[ \vec{L} \times \vec{B} = \hat{i} \left( -0.8 \times 10^{-3} \right) + 0 + 0. \] Thus: \[ \vec{L} \times \vec{B} = -(0.8 \times 10^{-3}) \, \hat{i}. \]
Step 2: Calculate $\vec{F$:}
\[ \vec{F} = I (\vec{L} \times \vec{B}) = 1.0 \cdot -(0.8 \times 10^{-3}) \, \hat{i}. \] Simplify: \[ \vec{F} = -0.8 \, \text{mN} \, \hat{i}. \] Thus, the force on the wire is: \[ \boxed{-(0.8 \, \text{mN}) \, \hat{i}}. \]