Question

Two infinite wires are placed at \( x = -2 \, \text{cm} \) and \( x = 1 \, \text{cm} \) along the y-axis, each carrying current upward. An electron is fired from the origin at \( 45^\circ \) to the x-axis with speed \( U \). The force on the electron at the moment of launch is:
(\( B_0 \) is the field at origin due to the wire at \( x = 1 \, \text{cm} \))

• A. \( \frac{-eUB_0}{2\sqrt{2}} (\hat{i} - \hat{j}) \)
• B. \( \frac{-eUB_0}{2} (\hat{i} - \hat{j}) \)
• C. \( \frac{-eUB_0}{\sqrt{2}} (\hat{i} - \hat{j}) \)
• D. \( -eUB_0 (\hat{i} - \hat{j}) \)

Hint:

Use \( \vec{F} = q (\vec{v} \times \vec{B}) \) and proper vector cross product for magnetic force.

Answer 1

The Correct Option is A

Magnetic field at origin is sum of fields due to both wires. Since one is at 1 cm and the other at -2 cm, their field magnitudes are in the ratio \( 2:1 \). So net magnetic field is: \[ B_{\text{net}} = B_0 - \frac{1}{2} B_0 = \frac{B_0}{2} \] Velocity vector at \( 45^\circ \): \( \vec{v} = \frac{U}{\sqrt{2}}(\hat{i} + \hat{j}) \)
Force: \( \vec{F} = -e (\vec{v} \times \vec{B}) \)
\( \vec{B} = \frac{B_0}{2} \hat{k} \Rightarrow \vec{F} = -e \cdot \frac{U}{\sqrt{2}} (\hat{i} + \hat{j}) \times \frac{B_0}{2} \hat{k} \)
Cross product: \( (\hat{i} + \hat{j}) \times \hat{k} = \hat{i} \times \hat{k} + \hat{j} \times \hat{k} = -\hat{j} + \hat{i} \Rightarrow \vec{F} = \frac{-eUB_0}{2\sqrt{2}} (\hat{i} - \hat{j}) \)