Question

A straight wire carrying a current of \( 2\sqrt{2} \, \text{A} \) is making an angle of \( 45^\circ \) with the direction of a uniform magnetic field of \( 3 \, \text{T} \). The force per unit length on the wire due to the magnetic field is:

• A. \( 4 \, \text{Nm}^{-1} \)
• B. \( 8 \, \text{Nm}^{-1} \)
• C. \( 6 \, \text{Nm}^{-1} \)
• D. \( 3 \, \text{Nm}^{-1} \)

Hint:

For magnetic force calculations, remember that the force per unit length is directly proportional to the current, magnetic field strength, and the sine of the angle between the field and the current direction.

Answer 1

The Correct Option is C

Step 1: Magnetic Force Per Unit Length Formula The force per unit length on a current-carrying conductor in a magnetic field is given by: \[ \frac{F}{L} = B I \sin \theta \] Where: - \( B = 3 \, \text{T} \) (magnetic field strength), - \( I = 2\sqrt{2} \, \text{A} \) (current), - \( \theta = 45^\circ \) (angle between the current and magnetic field). \vspace{0.5cm} 

Step 2: Substitute the Given Values Substitute the values into the formula: \[ \frac{F}{L} = 3 \times 2\sqrt{2} \times \sin 45^\circ \] Since \( \sin 45^\circ = \frac{\sqrt{2}}{2} \), we get: \[ \frac{F}{L} = 3 \times 2\sqrt{2} \times \frac{\sqrt{2}}{2} \] Simplifying the expression: \[ \frac{F}{L} = 3 \times 2 = 6 \, \text{Nm}^{-1} \] Thus, the correct answer is: \[ \mathbf{6 \, \text{Nm}^{-1}} \]