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 a wire segment due to a magnetic field, we use the formula for the magnetic force on a current-carrying wire:

\( \mathbf{F} = I \mathbf{L} \times \mathbf{B} \)

Where:

\( I \) is the current (0.5 A), \( \mathbf{L} \) is the length vector of the wire, and \( \mathbf{B} \) is the magnetic field vector.

The wire lies along the x-axis, and is 1 cm long, so its length vector is:

\( \mathbf{L} = (0.01 \, \text{m}) \hat{i} \)

Given the magnetic field vector:

\( \mathbf{B} = (0.4 \, \text{mT}) \hat{j} + (0.6 \, \text{mT}) \hat{k} \)

Convert milliteslas to teslas:

\( \mathbf{B} = (0.4 \times 10^{-3} \, \text{T}) \hat{j} + (0.6 \times 10^{-3} \, \text{T}) \hat{k} \)

Now, compute the cross product \( \mathbf{L} \times \mathbf{B} \):

\( \mathbf{L} \times \mathbf{B} = (0.01 \hat{i}) \times (0.4 \times 10^{-3} \hat{j} + 0.6 \times 10^{-3} \hat{k}) \)

Using the cross product rule:

• \( \hat{i} \times \hat{j} = \hat{k} \)
• \( \hat{i} \times \hat{k} = -\hat{j} \)

Calculate:

\( \mathbf{L} \times \mathbf{B} = 0.01(0.4 \times 10^{-3}) \hat{k} + 0.01(-0.6 \times 10^{-3}) (-\hat{j}) \)

\( = (0.4 \times 10^{-5}) \hat{k} - (0.6 \times 10^{-5}) \hat{j} \)

\( = (-0.6 \times 10^{-5}) \hat{j} + (0.4 \times 10^{-5}) \hat{k} \)

The magnitude of current \( I = 0.5 \, \text{A} \), so:

\( \mathbf{F} = 0.5((-0.6 \times 10^{-5}) \hat{j} + (0.4 \times 10^{-5}) \hat{k}) \)

\( = (-0.3 \times 10^{-5} \, \text{N}) \hat{j} + (0.2 \times 10^{-5} \, \text{N}) \hat{k} \)

Convert to micro Newtons (\( 1 \, \text{N} = 10^6 \, \mu\text{N} \)):

\( = (-3 \hat{j} + 2 \hat{k}) \, \mu\text{N} \)

Therefore, the force on the segment is: \((-3\hat{j} + 2\hat{k}) \, \mu\text{N}\).