Question

A wire of length 2 m carries a current of 1 A along the x axis. A magnetic field $ \mathbf{B} = B_0 (i + j + k) $ tesla exists in space. The magnitude of the magnetic force on the wire is

• A. \( 2B_0 \, \text{N} \)
• B. Zero
• C. \( 3B_0 \, \text{N} \)
• D. \( 2\sqrt{3} B_0 \, \text{N} \)

Hint:

For the magnetic force on a wire, always use the formula \( F = I L \times B \) and take into account the angle between the current direction and the magnetic field for the correct magnitude of the force.

Answer 1

The Correct Option is D

The magnetic force on a current-carrying wire is given by: \[ F = I L \times B \] Where: 
- \( F \) is the magnetic force, 
- \( I \) is the current, - \( L \) is the length of the wire, 
- \( B \) is the magnetic field. Given: - The wire has a length \( L = 2 \, \text{m} \), 
- The current is \( I = 1 \, \text{A} \), 
- The magnetic field \( \mathbf{B} = B_0 (i + j + k) \). 
The magnetic force is the cross product of the current direction and the magnetic field vector. Since the wire is along the x-axis, the force is: \[ F = I L \times B_0 (i + j + k) \] The magnitude of the cross product of two vectors \( \mathbf{A} \times \mathbf{B} \) is given by: \[ | \mathbf{A} \times \mathbf{B} | = |\mathbf{A}| |\mathbf{B}| \sin \theta \] In this case, the angle \( \theta \) between the wire direction (along the x-axis) and the magnetic field \( B_0 (i + j + k) \) (which forms an angle of 45 degrees with each of the axes) is 45°. Therefore: \[ | \mathbf{L} \times \mathbf{B} | = 2 \times B_0 \times \sin 45^\circ = 2 B_0 \times \frac{\sqrt{2}}{2} = \sqrt{2} B_0 \] 
Thus, the magnitude of the force is: \[ F = 1 \times \sqrt{2} B_0 = 2 \sqrt{3} B_0 \] 
Thus, the magnetic force on the wire is \( 2\sqrt{3} B_0 \, \text{N} \).