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Magnetic Effects of Electric Current - Direct current

Grade 10CBSE

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

🔑Concepts

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Direct Current (DCDC) is an electric current that flows in a single, constant direction. Unlike Alternating Current (ACAC), its polarity does not change over time.

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The frequency of DCDC is 0 Hz0 \text{ Hz} because the current does not complete any cycles or oscillations.

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Sources of DCDC include electrochemical cells, lead-acid batteries, and DCDC generators (dynamos).

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A straight current-carrying conductor produces a magnetic field (BB) in the form of concentric circles. The magnitude of the magnetic field is B∝IB \propto I and B∝1rB \propto \frac{1}{r}, where II is current and rr is the distance from the wire.

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The Right-Hand Thumb Rule states that if the thumb points in the direction of current (II), the curled fingers give the direction of the magnetic field lines (BB).

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A solenoid is a coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder. The magnetic field inside a long solenoid is uniform and parallel to its axis.

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The force (FF) experienced by a current-carrying conductor in a magnetic field is determined by Fleming's Left-Hand Rule, where the thumb is Force, the index finger is Magnetic Field (BB), and the middle finger is Current (II).

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The Electric Motor is a device that converts electrical energy (DCDC) into mechanical energy, utilizing the torque produced by the magnetic force on a current loop.

📐Formulae

B∝IB \propto I

B∝1rB \propto \frac{1}{r}

F=BIlsin⁡θF = BIl \sin \theta

f=0 Hzf = 0 \text{ Hz}

💡Examples

Problem 1:

A constant current of 2 A2 \text{ A} flows through a straight wire. If the distance from the wire is doubled, what happens to the magnitude of the magnetic field (BB)?

Solution:

The magnetic field will become half of its original value.

Explanation:

The magnetic field produced by a straight wire is inversely proportional to the distance from the wire (B∝1rB \propto \frac{1}{r}). If rr becomes 2r2r, then BB becomes B2\frac{B}{2}.

Problem 2:

An electron enters a magnetic field (BB) at a right angle, moving from North to South. The magnetic field is directed from West to East. Determine the direction of the force (FF) acting on the electron.

Solution:

The force acts vertically upwards (out of the plane).

Explanation:

Since the electron moves North to South, the conventional current (II) is South to North. Using Fleming's Left-Hand Rule: Forefinger (Field) points East, Middle finger (Current) points North. The Thumb then points upwards, indicating the direction of Force (FF).

Problem 3:

Calculate the force acting on a 0.5 m0.5 \text{ m} wire carrying a DCDC of 4 A4 \text{ A} placed perpendicular to a magnetic field of 0.1 T0.1 \text{ T}.

Solution:

F=0.1×4×0.5=0.2 NF = 0.1 \times 4 \times 0.5 = 0.2 \text{ N}

Explanation:

Using the formula F=BIlsin⁡θF = BIl \sin \theta, where θ=90∘\theta = 90^{\circ} (so sin⁡90∘=1\sin 90^{\circ} = 1), we substitute the values: F=(0.1 T)(4 A)(0.5 m)=0.2 NewtonsF = (0.1 \text{ T})(4 \text{ A})(0.5 \text{ m}) = 0.2 \text{ Newtons}.