Force on a Current-Carrying Wire - Detailed Solution

Force on a Current-Carrying Wire

Question: A straight wire of length \( L = 0.5 \) meter carrying a current \( I = 1.2 \) ampere is placed in a uniform magnetic field of induction \( B = 2 \) tesla. The magnetic field is perpendicular to the length of the wire. What is the force on the wire?

(a) \( 2.4\,\text{N} \)
(b) \( 1.2\,\text{N} \)
(c) \( 3.0\,\text{N} \)
(d) \( 2.0\,\text{N} \)

Detailed Step-by-Step Explanation

Step 1: Write Down the Relevant Formula
The magnetic force on a current-carrying wire is given by: \[ F = I L B \sin \theta, \] where \( \theta \) is the angle between the wire and the magnetic field.

Step 2: Substitute the Given Values
Since the magnetic field is perpendicular to the wire, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \). Thus, the force becomes: \[ F = I L B. \]

Step 3: Perform the Calculation
Substitute \( I = 1.2\,\text{A} \), \( L = 0.5\,\text{m} \), and \( B = 2\,\text{T} \) into the formula: \[ F = 1.2 \times 0.5 \times 2. \] Calculate: \[ 1.2 \times 0.5 = 0.6, \quad \text{and} \quad 0.6 \times 2 = 1.2. \]

Final Answer: The force on the wire is \( 1.2\,\text{N} \), which corresponds to option (b).