Question

A long straight current-carrying wire is placed in a uniform magnetic field of strength \( B = 0.5 \, \text{T} \). If the current in the wire is \( I = 2 \, \text{A} \) and the wire makes an angle of \( 30^\circ \) with the magnetic field, find the force per unit length on the wire.

• A. \( 0.5 \, \text{N/m} \)
• B. \( 2.0 \, \text{N/m} \)
• C. \( 1 \, \text{N/m} \)
• D. \( 3.0 \, \text{N/m} \)

Hint:

When calculating the force on a current-carrying wire in a magnetic field, remember that the force depends on the angle between the wire and the magnetic field. If the wire is parallel to the field, the force will be zero.

Answer 1

The Correct Option is A

Given:

• Magnetic field strength: \( B = 0.5 \, \text{T} \)
• Current in the wire: \( I = 2 \, \text{A} \)
• Angle with magnetic field: \( \theta = 30^\circ \)

Step 1: Use the formula for the force on a current-carrying wire in a magnetic field

The formula for the force per unit length on a current-carrying wire in a magnetic field is: \[ \frac{F}{L} = B I \sin \theta \] where: - \( \frac{F}{L} \) is the force per unit length, - \( B \) is the magnetic field strength, - \( I \) is the current, - \( \theta \) is the angle between the magnetic field and the current.

Step 2: Substitute the given values into the formula

\[ \frac{F}{L} = (0.5 \, \text{T}) \times (2 \, \text{A}) \times \sin(30^\circ) \] Since \( \sin(30^\circ) = 0.5 \), we get: \[ \frac{F}{L} = 0.5 \times 2 \times 0.5 = 0.5 \, \text{N/m} \]

✅ Final Answer:

The force per unit length on the wire is \( \boxed{0.5 \, \text{N/m}} \).