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Magnetic Effects of Electric Current - Fleming’s Left Hand Rule

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Fleming’s Left Hand Rule states that if we stretch the thumb, forefinger, and middle finger of the left hand such that they are mutually perpendicular to each other, each finger represents a specific vector direction.

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The Forefinger represents the direction of the Magnetic Field (BB).

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The Middle Finger represents the direction of the Current (II).

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The Thumb points in the direction of the Force (FF) or the Motion acting on the conductor.

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This rule is primarily used to determine the direction of force in an electric motor.

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The force experienced by a current-carrying conductor is maximum when the angle between the current and the magnetic field is θ=90∘\theta = 90^{\circ}.

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If the conductor is placed parallel to the magnetic field (θ=0∘\theta = 0^{\circ}), the force experienced is zero (F=0F = 0).

📐Formulae

F=BIlsin⁡θF = BIl \sin \theta

Fmax=BIl (when θ=90∘)F_{max} = BIl \text{ (when } \theta = 90^{\circ} \text{)}

F∝IF \propto I

F∝BF \propto B

💡Examples

Problem 1:

An electron enters a magnetic field at right angles to it. The direction of the magnetic field is from North to South, and the electron is moving from West to East. Find the direction of the force acting on the electron.

Solution:

The direction of the force is vertically upwards, out of the plane of the paper.

Explanation:

  1. The electron moves from West to East, so the conventional current (II) is directed from East to West.
  2. The magnetic field (BB) is North to South.
  3. Using Fleming’s Left Hand Rule: point the Forefinger South (Field) and the Middle finger West (Current). The Thumb points vertically upwards.

Problem 2:

A current-carrying conductor of length 1.5 m1.5\text{ m} is placed perpendicular to a magnetic field of 2 T2\text{ T}. If the current flowing through it is 5 A5\text{ A}, calculate the force acting on it.

Solution:

F=15 NF = 15\text{ N}

Explanation:

Given: l=1.5 ml = 1.5\text{ m}, B=2 TB = 2\text{ T}, I=5 AI = 5\text{ A}, and θ=90∘\theta = 90^{\circ}. Using the formula F=BIlsin⁡90∘F = BIl \sin 90^{\circ}, we get F=2×5×1.5×1F = 2 \times 5 \times 1.5 \times 1, which equals 15 N15\text{ N}.

Fleming’s Left Hand Rule - Revision Notes & Key Formulas | CBSE Class 10 Science