recall and use the equation F = BIL sin θ, with directions as interpreted by Fleming’s left-hand rule

1. Introduction

When a conductor carrying an electric current is placed in a magnetic field, the magnetic field exerts a mechanical force on the conductor. This phenomenon is fundamental to many devices such as motors, galvanometers and loudspeakers.

2. The Fundamental Equation

The magnitude of the force F on a straight conductor of length L carrying a current I in a uniform magnetic field of flux density B is given by

\$\$

F = B I L \sin\theta

\$\$

where θ is the angle between the direction of the current (the line of the conductor) and the direction of the magnetic field.

2.1 Meaning of each symbol

• F – magnitude of the magnetic force (N)
• B – magnetic flux density (tesla, T)
• I – current (ampere, A)
• L – length of the conductor that lies within the magnetic field (m)
• θ – angle between the direction of I (or L) and B

2.2 \cdot ector form

In vector notation the force is expressed as

\$\$

\mathbf{F} = I \, \mathbf{L} \times \mathbf{B}

\$\$

where 𝑳 is a vector of magnitude L in the direction of conventional current, and “×” denotes the vector cross‑product.

3. Determining the Direction – Fleming’s Left‑Hand Rule

Fleming’s left‑hand rule provides a quick way to find the direction of the force on a current‑carrying conductor.

1. Stretch the thumb, fore‑finger and middle finger of the left hand so that they are mutually perpendicular.
2. Point the fore‑finger in the direction of the magnetic field B (from north to south).
3. Point the middle finger in the direction of the conventional current I (positive to negative).
4. The thumb then points in the direction of the force F on the conductor.

3.1 Using the rule with the equation

If the conductor is not perpendicular to the field (θ ≠ 90°), the component of the field that is effective is B sin θ. The direction given by the left‑hand rule still applies to this effective component.

4. Worked Example

Problem: A straight horizontal wire of length 0.25 m carries a current of 3.0 A to the east. It is placed in a uniform magnetic field of 0.40 T directed vertically upwards. Find the magnitude and direction of the force on the wire.

Solution:

1. Identify the vectors:

   • Current direction (middle finger) – east.
   • Magnetic field direction (fore‑finger) – up.

2. Apply Fleming’s left‑hand rule: thumb points to the south. Hence the force is directed towards the south.
3. Since the wire is perpendicular to the field, θ = 90° and sin θ = 1.
4. Calculate the magnitude:

   \$F = B I L \sin\theta = (0.40\ \text{T})(3.0\ \text{A})(0.25\ \text{m})(1) = 0.30\ \text{N}\$

The force on the wire is 0.30 N towards the south.

5. Common Pitfalls

• Confusing conventional current direction with electron flow – always use conventional (positive to negative) for the left‑hand rule.
• Neglecting the sin θ factor when the conductor is not perpendicular to the field.
• Using the right‑hand rule (which applies to the force on a moving charge) instead of Fleming’s left‑hand rule for a current‑carrying conductor.
• For a curved conductor, treat each infinitesimal segment dL separately and integrate: \$\displaystyle \mathbf{F} = I\int \mathrm{d}\mathbf{L}\times\mathbf{B}\$.

6. Practice Questions

7. Summary Checklist

• Write down the known quantities and identify the direction of B and I.
• Determine the angle θ between the conductor (or current direction) and the magnetic field.
• Use \$F = B I L \sin\theta\$ to calculate the magnitude.
• Apply Fleming’s left‑hand rule to obtain the direction of the force.
• Check units (T·A·m = N) and sign conventions.