Question

A straight conductor is carrying a current of 2 A in the \( +x \) direction along it. A uniform magnetic field \( \vec{B} = (0.6 \hat{j} + 0.8 \hat{k}) \) T is switched on in the region. The force acting on 10 cm length of the conductor is:

• A. \( (0.12 \hat{j} - 0.16 \hat{k}) \, \text{N} \)
• B. \( (-0.16 \hat{j} + 0.12 \hat{k}) \, \text{N} \)
• C. \( (-0.12 \hat{j} + 0.16 \hat{k}) \, \text{N} \)
• D. \( (0.16 \hat{j} - 0.12 \hat{k}) \, \text{N} \)

Hint:

The direction of the force can be determined using the right-hand rule for the cross product. The direction of the magnetic force on the conductor is perpendicular to both the current and the magnetic field.

Answer 1

The Correct Option is B

The magnetic force on a current-carrying wire is quantified by the equation: \[ \vec{F} = I \, \ell \, \vec{B} \times \hat{l} \] where: - \( I \) denotes the current in the conductor, given as 2 A. - \( \ell \) represents the length of the conductor, specified as 0.1 m. - \( \vec{B} \) signifies the magnetic field. - \( \hat{l} \) is the unit vector indicating the direction of the current, which is \( \hat{i} \) along the \( +x \)-direction. Given parameters: - \( I = 2 \, \text{A} \) - \( \ell = 0.1 \, \text{m} \) - \( \vec{B} = (0.6 \hat{j} + 0.8 \hat{k}) \, \text{T} \) - \( \hat{l} = \hat{i} \) The cross product \( \vec{B} \times \hat{l} \) is computed as follows: \[ \vec{B} \times \hat{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0.6 & 0.8 & 0 \end{vmatrix} \] \[ = \hat{i} \left( 0 \cdot 0 - 0.8 \cdot 0 \right) - \hat{j} \left( 0 \cdot 0 - 0.6 \cdot 0 \right) + \hat{k} \left( 1 \cdot 0.8 - 0.6 \cdot 0 \right) \] \[ = 0 \hat{i} - 0 \hat{j} + 0.8 \hat{k} \] Therefore, \( \vec{B} \times \hat{l} = 0 \hat{i} + 0 \hat{j} + 0.8 \hat{k} \). The resulting force is: \[ \vec{F} = I \, \ell \, \vec{B} \times \hat{l} = 2 \times 0.1 \times (0 \hat{i} + 0 \hat{j} + 0.8 \hat{k}) \] \[ \vec{F} = (0 \hat{i} + 0.16 \hat{j} + 0.12 \hat{k}) \, \text{N} \] The final calculated force is \({(-0.16 \hat{j} + 0.12 \hat{k}) \, \text{N}} \).