Question

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

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

Hint:

Remember, the force on a current-carrying conductor in a magnetic field is given by \( F = I \ell (\vec{B} \times \hat{l}) \). The direction of the force is given by the right-hand rule.

Answer 1

The Correct Option is D

To find the force acting on the wire segment, we use the formula for the magnetic force on a current-carrying conductor: \(\vec{F} = I (\vec{L} \times \vec{B})\), where \(I\) is the current, \(\vec{L}\) is the length vector of the wire, and \(\vec{B}\) is the magnetic field. Given values are \(I = 0.5 \, \text{A}\), \(\vec{L} = 1 \, \text{cm} = 0.01 \, \text{m} \, \hat{i}\), and \(\vec{B} = (0.4 \, \text{mT}) \hat{j} + (0.6 \, \text{mT}) \hat{k}\). First, convert the magnetic field units: \(1 \, \text{mT} = 10^{-3} \, \text{T}\), so \(\vec{B} = (0.4 \times 10^{-3}) \hat{j} + (0.6 \times 10^{-3}) \hat{k} \, \text{T}\).

Next, calculate the cross product \(\vec{L} \times \vec{B}\): 

\[\vec{L} \times \vec{B} = (0.01 \, \hat{i}) \times \left((0.4 \times 10^{-3}) \hat{j} + (0.6 \times 10^{-3}) \hat{k}\right)\]

\[= 0.01 \times 0.4 \times 10^{-3} \, (\hat{i} \times \hat{j}) + 0.01 \times 0.6 \times 10^{-3} \, (\hat{i} \times \hat{k})\]

\[= 0.4 \times 10^{-5} \, \hat{k} - 0.6 \times 10^{-5} \, \hat{j}\] (using \(\hat{i} \times \hat{j} = \hat{k}\) and \(\hat{i} \times \hat{k} = -\hat{j}\))

Thus, \(\vec{L} \times \vec{B} = (0 \hat{i} - 0.6 \times 10^{-5} \hat{j} + 0.4 \times 10^{-5} \hat{k})\).

Now, multiply by current \(I\):

\[\vec{F} = 0.5 \times (0 \hat{i} - 0.6 \times 10^{-5} \hat{j} + 0.4 \times 10^{-5} \hat{k})\]

\[= (0 \times 0.5) \hat{i} - (0.6 \times 10^{-5} \times 0.5) \hat{j} + (0.4 \times 10^{-5} \times 0.5) \hat{k}\]

\[= -0.3 \times 10^{-5} \hat{j} + 0.2 \times 10^{-5} \hat{k}\]

Convert \(\vec{F}\) to \(\text{mN}\):

\[= -3 \times 10^{-6} \hat{j} + 2 \times 10^{-6} \hat{k} \, \text{N}\]

This can be expressed in \(\text{mN}\) as:

\[= -3 \hat{j} \, \mu\text{N} + 6 \hat{k} \, \mu\text{N} = (-3 \hat{i} + 6 \hat{k}) \, \mu\text{N}\]

The correct force is \( (-3 \hat{i} + 6 \hat{k}) \, \text{mN} \).