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Physics: Electricity and Magnetism - Magnets, Electromagnets, and Electromagnetic Forces

Grade 8IB

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

🔑Concepts

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Magnetic Poles: All magnets have a North (NN) and a South (SS) pole. Like poles repel, and opposite poles attract.

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Magnetic Field (BB): The region around a magnet where magnetic forces are exerted. Field lines are drawn from the North pole to the South pole.

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Magnetic Materials: Materials like Iron, Nickel, and Cobalt are ferromagnetic and can be magnetized.

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Electromagnetism: An electric current (II) flowing through a conductor produces a magnetic field around it. The direction of the field can be determined using the Right-Hand Grip Rule.

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Solenoids: A coil of wire acting as a magnet when carrying an electric current. The magnetic field inside a solenoid is strong and uniform.

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Electromagnets: Temporary magnets made by winding a coil around a soft iron core. Their strength depends on the current (II), the number of turns (NN), and the core material.

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The Motor Effect: When a current-carrying wire is placed in an external magnetic field, it experiences a force (FF).

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Fleming's Left-Hand Rule: Used to find the direction of the force. The Thumb represents Force (FF), the First Finger represents Magnetic Field (BB), and the Second Finger represents Current (II).

📐Formulae

F=BIlF = BIl

V=IRV = IR

B∝IB \propto I

B∝NB \propto N

💡Examples

Problem 1:

A wire of length l=0.2 ml = 0.2\text{ m} carries a current of I=5 AI = 5\text{ A} at right angles to a magnetic field with a strength of B=0.04 TB = 0.04\text{ T}. Calculate the magnetic force (FF) acting on the wire.

Solution:

F=BIlF = BIl F=0.04×5×0.2F = 0.04 \times 5 \times 0.2 F=0.04 NF = 0.04\text{ N}

Explanation:

We use the formula for the motor effect where the force is the product of magnetic flux density, current, and the length of the wire within the field.

Problem 2:

An electromagnet is constructed with a solenoid. If the current flowing through the circuit is doubled from II to 2I2I and the number of turns in the coil is tripled from NN to 3N3N, by what factor does the magnetic field strength (BB) increase?

Solution:

Bnew∝(2I)×(3N)B_{new} \propto (2I) \times (3N) Bnew∝6(I×N)B_{new} \propto 6(I \times N) Factor=6\text{Factor} = 6

Explanation:

The magnetic field strength of an electromagnet is directly proportional to both the current and the number of turns. Multiplying the current by 22 and the turns by 33 results in a total increase of 2×3=62 \times 3 = 6 times the original strength.