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Magnetic Effects of Electric Current - 12.312.3 12.312.3 FORCE ON A CURRENTFORCE ON A CURRENTFORCE ON A CURRENTFORCE ON A CURRENTFORCE ON A CURRENT ---

Grade 10CBSE

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

🔑Concepts

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André-Marie Ampère suggested that if a current-carrying conductor produces a magnetic field and exerts a force on a magnet, then a magnet must also exert an equal and opposite force on the current-carrying conductor.

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The displacement of the conductor is maximum (and the magnitude of the force is highest) when the direction of the current is at right angles (90∘90^\circ) to the direction of the magnetic field.

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Fleming's Left-Hand Rule: Stretch the thumb, forefinger, and middle finger of your left hand such that they are mutually perpendicular. If the first finger (forefinger) points in the direction of the magnetic field and the second finger (middle finger) points in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

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The direction of the force on the conductor is reversed if the direction of current is reversed or if the direction of the magnetic field is reversed.

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Devices that use current-carrying conductors and magnetic fields include electric motors, electric generators, loudspeakers, and microphones.

📐Formulae

F=BIlsin⁡θF = B I l \sin \theta

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

Fmin=0(when θ=0∘ or 180∘)F_{min} = 0 \quad (\text{when } \theta = 0^\circ \text{ or } 180^\circ)

💡Examples

Problem 1:

An electron enters a magnetic field at right angles to it. The magnetic field is directed from left to right. The electron is moving vertically upwards. What is the direction of the force acting on the electron?

Solution:

The direction of force is into the page (perpendicularly downwards into the surface).

Explanation:

According to Fleming's Left-Hand Rule, we must first identify the direction of the current. Since electrons are negatively charged and moving upwards, the conventional current II is directed downwards. The magnetic field BB is directed towards the right. Aligning the middle finger (current) downwards and the forefinger (field) to the right, the thumb (force) points into the page.

Problem 2:

A current-carrying wire of length l=0.5 ml = 0.5\text{ m} is placed perpendicular to a magnetic field of B=2 TB = 2\text{ T}. If the current flowing through the wire is I=1.5 AI = 1.5\text{ A}, calculate the force acting on the wire.

Solution:

F=1.5 NF = 1.5\text{ N}

Explanation:

Using the formula for force F=BIlsin⁡θF = BIl \sin \theta. Since the wire is perpendicular, θ=90∘\theta = 90^\circ and sin⁡90∘=1\sin 90^\circ = 1. F=2×1.5×0.5F=3.0×0.5F=1.5 N\begin{array}{r} F = 2 \times 1.5 \times 0.5 \\ F = 3.0 \times 0.5 \\ \hline F = 1.5\text{ N} \end{array}