Question

The force per unit length on a straight wire carrying current of 8 A making an angle of \(30^\circ\) with a uniform magnetic field of 0.15 T is

• A. \( 1.2 \, \text{N m}^{-1} \)
• B. \( 1.02 \, \text{N m}^{-1} \)
• C. \( 0.6 \, \text{N m}^{-1} \)
• D. \( 2.4 \, \text{N m}^{-1} \)

Hint:

To find magnetic force on a current-carrying wire, use \( F = I L B \sin\theta \). For force per unit length, use \( \frac{F}{L} = I B \sin\theta \). Always check the angle between current and magnetic field.

Answer 1

The Correct Option is C

Step 1: Use the formula for magnetic force on a current-carrying wire.
The magnetic force per unit length on a straight wire is given by: \[ \frac{F}{L} = I B \sin\theta \] Where: \( I = 8 \, \text{A} \) (current) 
\( B = 0.15 \, \text{T} \) (magnetic field) 
\( \theta = 30^\circ \) 
Step 2: Substitute the values.
\[ \frac{F}{L} = 8 \times 0.15 \times \sin(30^\circ) \] \[ \frac{F}{L} = 8 \times 0.15 \times 0.5 = 0.6 \, \text{N m}^{-1} \] Step 3: Select the correct option.
The calculated magnetic force per unit length is \( 0.6 \, \text{N m}^{-1} \), which is option (3).