Question

A 1 cm segment of a wire lying along the x-axis carries a current of 0.5 A along the \( +x \)-direction. A magnetic field \( \mathbf{B} = (0.4 \, \text{mT}) \hat{j + (0.6 \, \text{mT}) \hat{k} \) is switched on in the region. The force acting on the segment is:}

• A. \( (2\hat{j} + 3\hat{k}) \, \text{mN} \)
• B. \( (-3\hat{j} + 2\hat{k}) \, \mu\text{N} \)
• C. \( (6\hat{j} + 4\hat{k}) \, \text{mN} \)
• D. \( (-4\hat{j} + 6\hat{k}) \, \mu\text{N} \)

Hint:

The force on a current-carrying conductor in a magnetic field is calculated using \( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \).

Answer 1

The Correct Option is B

To determine the force acting on the segment of the wire, we can use the Lorentz force law for a current-carrying wire in a magnetic field:

F = I L × B

where:

• I = 0.5A is the current,
• L = 1 cm î = 0.01 m î is the length of the wire segment,
• B = (0.4 mT) ĵ + (0.6 mT) ĸ = (0.4 × 10⁻³ T) ĵ + (0.6 × 10⁻³ T) ĸ is the magnetic field.

Now, compute the cross product L × B:

L × B =

\[ \hat{i} \quad L × B = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0.01 & 0 & 0 \\ 0 & 0.4 \times 10^{-3} & 0.6 \times 10^{-3} \end{matrix} \right| \]

Expanding the determinant:

L × B = \[ \hat{i}(0 \cdot 0.6 \times 10^{-3} - 0 \cdot 0.4 \times 10^{-3}) - \hat{j}(0.01 \cdot 0.6 \times 10^{-3} - 0 \cdot 0) + \hat{k}(0.01 \cdot 0.4 \times 10^{-3} - 0 \cdot 0) \]

L × B = \[ - \hat{j}(6 \times 10^{-6}) + \hat{k}(4 \times 10^{-6}) \]

Now, multiply by the current I = 0.5A:

F = 0.5 × (−6 × 10⁻⁶) ĵ + 0.5 × (4 × 10⁻⁶) ĸ

F = −3 × 10⁻⁶ ĵ + 2 × 10⁻⁶ ĸ

Thus, the correct answer is:

F = (−3 ĵ + 2 ĸ) µN