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Magnetic Effects of Electric Current - Force on current carrying conductor

Grade 10CBSE

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

🔑Concepts

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A current-carrying conductor placed in a magnetic field experiences a mechanical force due to the interaction between the external magnetic field and the magnetic field produced by the current in the conductor.

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This phenomenon was first suggested by André-Marie Ampère, who stated that if a magnetic field exerts a force on a current-carrying conductor, then the conductor must also exert an equal and opposite force on the magnet.

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The direction of the force is always perpendicular to both the direction of the current (II) and the direction of the magnetic field (BB).

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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 forefinger points in the direction of the magnetic field (BB) and the middle finger points in the direction of current (II), then the thumb points in the direction of the force (FF) or motion.

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

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The force is zero when the direction of current is parallel or anti-parallel to the direction of the magnetic field (heta=0∘ heta = 0^\circ or 180∘180^\circ).

📐Formulae

F=BIlsin⁡θF = B I l \sin \theta

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

F=0 (when θ=0∘ or 180∘)F = 0 \text{ (when } \theta = 0^\circ \text{ or } 180^\circ\text{)}

💡Examples

Problem 1:

A wire of length l=0.2 ml = 0.2\text{ m} carrying a current of I=5 AI = 5\text{ A} is placed perpendicular to a uniform magnetic field of B=0.1 TB = 0.1\text{ T}. Calculate the magnitude of the force acting on the wire.

Solution:

Given: l=0.2 ml = 0.2\text{ m}, I=5 AI = 5\text{ A}, B=0.1 TB = 0.1\text{ T}, θ=90∘\theta = 90^\circ. Using the formula F=BIlsin⁡θF = B I l \sin \theta, we get F=0.1×5×0.2×sin⁡90∘F = 0.1 \times 5 \times 0.2 \times \sin 90^\circ. Since sin⁡90∘=1\sin 90^\circ = 1, F=0.1×5×0.2=0.1 NF = 0.1 \times 5 \times 0.2 = 0.1\text{ N}.

Explanation:

Since the wire is perpendicular to the magnetic field, the maximum force is exerted. The force is calculated by the product of magnetic field strength, current, and the length of the wire.

Problem 2:

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

Solution:

  1. Direction of electron flow: West to East. 2. Direction of conventional current (II): East to West (opposite to electron flow). 3. Direction of Magnetic Field (BB): North to South. 4. Applying Fleming's Left Hand Rule: Forefinger (Field) points South, Middle finger (Current) points West. The Thumb points vertically upwards.

Explanation:

The direction of force is determined by Fleming's Left Hand Rule. It is crucial to remember that the direction of current is taken as opposite to the direction of motion of negatively charged particles like electrons.

Force on current carrying conductor Revision - Class 10 Science CBSE