Question

A conductor of length 20 cm carrying a curent of 5A is placed at an angle of 30° to the extemal magnetic field of 0.5 T. The.force acting on it is

• A. 0.5 N
• B. 5N
• C. 2.5N
• D. 0.25N
• E. 0.125N

Answer 1

The Correct Option is D

Force on a Current-Carrying Conductor in a Magnetic Field 

A conductor of length 20 cm carrying a current of 5 A is placed at an angle of 30° to an external magnetic field of 0.5 T. We need to determine the force acting on it.

Step 1: Formula for Magnetic Force

The force (F) on a straight current-carrying conductor of length \(L\) carrying a current \(I\) placed in a magnetic field \(B\) at an angle \(\theta\) to the field is given by:

\(F = B I L \sin\theta\)

Where:

• \(F\) is the magnitude of the magnetic force.
• \(B\) is the magnitude of the magnetic field.
• \(I\) is the current.
• \(L\) is the length of the conductor.
• \(\theta\) is the angle between the conductor and the magnetic field.

Step 2: Convert Length to Meters

We are given the length in centimeters, so we need to convert it to meters:

\(L = 20 \text{ cm} = 0.2 \text{ m}\)

Step 3: Plug in the Values

Now, we plug in the given values into the formula:

\(F = (0.5 \text{ T}) (5 \text{ A}) (0.2 \text{ m}) \sin(30^\circ)\)

Step 4: Calculate the Force

We know that \(\sin(30^\circ) = 0.5\), so:

\(F = (0.5)(5)(0.2)(0.5) \text{ N}\)

\(F = 0.25 \text{ N}\)

Conclusion

The force acting on the conductor is 0.25 N.

Answer 2

The magnetic force on a current-carrying conductor is given by: \[ F = BIL \sin\theta \] Where: \( B = 0.5 \, \text{T} \) (magnetic field) \( I = 5 \, \text{A} \) (current) \( L = 20 \, \text{cm} = 0.2 \, \text{m} \) (length of conductor) \( \theta = 30^\circ \) (angle between conductor and magnetic field) Substitute into the formula: \[ F = 0.5 \times 5 \times 0.2 \times \sin 30^\circ = 0.5 \times 5 \times 0.2 \times \frac{1}{2} = 0.25 \, \text{N} \]