Question

The magnetic force on a current carrying wire placed in a uniform magnetic field if the wire is oriented perpendicular to the magnetic field is

• A. F=BIL
• B. \(F=\frac{B}{I}\)
• C. \(F=\frac{L}{BI}\)
• D. \(F=\frac{I}{BL}\)

Answer 1

The Correct Option is A

To solve the problem, we need to determine the magnetic force acting on a current-carrying wire placed perpendicular to a magnetic field.

1. Understanding the Formula:
The magnetic force $F$ on a current-carrying conductor placed in a uniform magnetic field is given by:

$ F = BIL \sin \theta $
where:
$B$ = magnetic field strength
$I$ = current
$L$ = length of the conductor in the magnetic field
$\theta$ = angle between the wire and the magnetic field

2. Given Condition:
The wire is oriented perpendicular to the magnetic field, so:
$ \theta = 90^\circ \Rightarrow \sin \theta = 1 $

3. Substituting in the Formula:
$ F = BIL \cdot \sin 90^\circ = BIL \cdot 1 = BIL $

Final Answer:
The magnetic force is $ F = BIL $.