Question

The magnetic force on a current carrying wire of length $l$, placed in a uniform magnetic field ($B$) if the wire is oriented perpendicular to magnetic field, is \dots

• A. 0
• B. $ILB$
• C. $2ILB$
• D. $\frac{ILB}{2}$

Hint:

The magnetic force becomes maximum when the conductor is perpendicular to the magnetic field because: \[ \sin 90^\circ = 1 \]

Answer 1

The Correct Option is B

Concept: A current carrying conductor placed inside a magnetic field experiences a magnetic force. This phenomenon is the working principle of electric motors. The magnitude of magnetic force acting on a conductor is given by: \[ F = BIl \sin\theta \] Where:
• $F$ = magnetic force
• $B$ = magnetic field strength
• $I$ = current in the conductor
• $l$ = length of conductor
• $\theta$ = angle between current direction and magnetic field

Step 1: Identify the angle. The conductor is perpendicular to the magnetic field. Therefore, \[ \theta = 90^\circ \]

Step 2: Use the trigonometric value. \[ \sin 90^\circ = 1 \]

Step 3: Substitute into the formula. \[ F = BIl \times 1 \] \[ F = BIl \] Thus, the magnetic force acting on the conductor is: \[ ILB \]