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Physics - Electromagnetism

Grade 10ICSE

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

🔑Concepts

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Oersted's Experiment: Demonstrated that a current-carrying conductor produces a magnetic field B⃗\vec{B} around it.

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Right Hand Thumb Rule: If the thumb points in the direction of current II, the curled fingers show the direction of magnetic field lines.

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Magnetic Field of a Solenoid: A long coil of insulated copper wire. The magnetic field inside is uniform and proportional to the current II and number of turns per unit length nn, given by B=μ0nIB = \mu_0 n I.

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Electromagnet vs. Permanent Magnet: An electromagnet's polarity can be reversed and its strength changed by varying II, whereas a permanent magnet has fixed properties.

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Force on a Conductor: A current-carrying wire of length ll in a magnetic field BB experiences a force F=BIlsin⁡θF = BIl \sin \theta. The force is maximum when θ=90∘\theta = 90^\circ.

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Fleming's Left Hand Rule: Used to find the direction of force (ThumB - Motion/Force, Forefinger - Field, Middle finger - Current).

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Electromagnetic Induction (EMI): The phenomenon of producing an induced e.m.f.e.m.f. by changing the magnetic flux ϕ\phi linked with a circuit.

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Faraday's Laws: 1) Change in flux induces e.m.f.e.m.f. 2) Magnitude of induced e.m.f.e.m.f. ε\varepsilon is proportional to the rate of change of magnetic flux.

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Lenz's Law: The direction of induced current is such that it opposes the change in magnetic flux that produced it (Conservation of Energy).

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Fleming's Right Hand Rule: Used for induced current (ThumB - Motion, Forefinger - Field, Middle finger - Induced Current).

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Transformer: A device used to change AC voltage based on mutual induction. In a step-up transformer, Ns>NpN_s > N_p, and in a step-down transformer, Ns<NpN_s < N_p.

📐Formulae

F=BIlsin⁡θF = BIl \sin \theta

ϕ=B⋅Acos⁡θ\phi = B \cdot A \cos \theta

ε=−NΔϕΔt\varepsilon = -N \frac{\Delta \phi}{\Delta t}

VsVp=NsNp=k\frac{V_s}{V_p} = \frac{N_s}{N_p} = k

IpIs=NsNp=VsVp\frac{I_p}{I_s} = \frac{N_s}{N_p} = \frac{V_s}{V_p}

Efficiency (η)=PoutPin×100=VsIsVpIp×100\text{Efficiency } (\eta) = \frac{P_{out}}{P_{in}} \times 100 = \frac{V_s I_s}{V_p I_p} \times 100

💡Examples

Problem 1:

A transformer has 500500 turns in the primary coil and 25002500 turns in the secondary coil. If the primary voltage is 220 V220\text{ V}, calculate the secondary voltage and identify the type of transformer.

Solution:

Given: Np=500N_p = 500, Ns=2500N_s = 2500, Vp=220 VV_p = 220\text{ V}. Using the transformer formula: VsVp=NsNp  ⟹  Vs=Vp×NsNp=220×2500500=220×5=1100 V\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = V_p \times \frac{N_s}{N_p} = 220 \times \frac{2500}{500} = 220 \times 5 = 1100\text{ V}.

Explanation:

Since Vs>VpV_s > V_p and Ns>NpN_s > N_p, this is a step-up transformer.

Problem 2:

A straight conductor of length 0.5 m0.5\text{ m} carrying a current of 2 A2\text{ A} is placed perpendicular to a magnetic field of 0.4 T0.4\text{ T}. Calculate the force acting on the conductor.

Solution:

Given: l=0.5 ml = 0.5\text{ m}, I=2 AI = 2\text{ A}, B=0.4 TB = 0.4\text{ T}, θ=90∘\theta = 90^\circ. Force F=BIlsin⁡90∘=0.4×2×0.5×1=0.4 NF = BIl \sin 90^\circ = 0.4 \times 2 \times 0.5 \times 1 = 0.4\text{ N}.

Explanation:

The force is calculated using the Lorentz force formula for a conductor. Since it is perpendicular, sin⁡90∘=1\sin 90^\circ = 1 gives the maximum force.

Electromagnetism - Revision Notes & Key Formulas | ICSE Class 10 Science