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 force on a current-carrying conductor in a magnetic field is given by the formula: \[ \vec{F} = I \, \ell \, \vec{B} \times \hat{l} \] where: - \( I \) is the current in the conductor (2 A), - \( \ell \) is the length of the conductor (0.1 m), - \( \vec{B} \) is the magnetic field, and - \( \hat{l} \) is the unit vector along the direction of the current, which is \( \hat{i} \) in the \( +x \)-direction. Given: - \( I = 2 \, \text{A} \), - \( \ell = 0.1 \, \text{m} \), - \( \vec{B} = (0.6 \hat{j} + 0.8 \hat{k}) \, \text{T} \), - \( \hat{l} = \hat{i} \). Now, calculate the cross product \( \vec{B} \times \hat{l} \): \[ \vec{B} \times \hat{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 0 & 0
0.6 & 0.8 & 0 \end{vmatrix} \] \[ = \hat{i} \left( 0 \cdot 0 - 0.8 \cdot 0 \right) - \hat{j} \left( 0 \cdot 0 - 0.6 \cdot 0 \right) + \hat{k} \left( 1 \cdot 0.8 - 0.6 \cdot 0 \right) \] \[ = 0 \hat{i} - 0 \hat{j} + 0.8 \hat{k} \] Thus, the cross product is \( \vec{B} \times \hat{l} = 0 \hat{i} + 0 \hat{j} + 0.8 \hat{k} \). Now, the force is: \[ \vec{F} = I \, \ell \, \vec{B} \times \hat{l} = 2 \times 0.1 \times (0 \hat{i} + 0 \hat{j} + 0.8 \hat{k}) \] \[ \vec{F} = (0 \hat{i} + 0.16 \hat{j} + 0.12 \hat{k}) \, \text{N} \] Thus, the correct answer is \({(-0.16 \hat{j} + 0.12 \hat{k}) \, \text{N}} \).