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Chapter 1: Electrostatics

Short Answer Questions (10)

1. Define electric flux and write its SI unit.

Electric flux is the measure of flow of electric field through a given area. It is defined as the dot product of electric field and area vector. Its SI unit is Nm²/C or Vm.

2. State Coulomb's law in vector form.

F⃗ = (k q₁ q₂)/r² · r̂ where F⃗ is force, q₁ and q₂ are charges, r is distance between them, r̂ is unit vector along the line joining charges, and k is Coulomb's constant (9×10⁹ Nm²/C²).

3. What is the principle of superposition in electrostatics?

The principle states that the net force on any charge due to a group of charges is the vector sum of all the individual forces exerted on it by each charge in the group, with all other charges being absent.

4. Differentiate between conductors and insulators.

Conductors allow free movement of electric charges (e.g., metals), while insulators resist charge movement (e.g., rubber, glass). Conductors have high conductivity and low resistivity, opposite for insulators.

5. What is an electric dipole? Give its moment expression.

An electric dipole is a pair of equal and opposite charges (+q and -q) separated by small distance (2a). Its dipole moment is p⃗ = q × 2a⃗ (from -q to +q), measured in Cm.

6. Explain why electric field inside a conductor is zero.

In electrostatic equilibrium, free electrons redistribute to cancel any external field. The charges arrange themselves on the surface until the net field inside becomes zero.

7. Define dielectric constant of a medium.

Dielectric constant (K or εᵣ) is the ratio of permittivity of medium (ε) to permittivity of free space (ε₀). It measures how much the medium reduces the electric field compared to vacuum.

8. What is Gauss's law? Write its mathematical form.

Gauss's law states that total electric flux through any closed surface equals 1/ε₀ times the net charge enclosed. ∮E⃗·dA⃗ = Q_enclosed/ε₀.

9. Why can't two electric field lines intersect?

Electric field lines cannot intersect because at any point, the electric field has only one direction. If lines crossed, it would imply two directions simultaneously, which is impossible.

10. What is the work done in moving a charge in an electrostatic field?

Work done is path independent and equals the change in potential energy: W = qΔV = q(V_final - V_initial). For a closed loop, net work done is zero.

Numerical Problems (7)

1. Calculate the force between two charges of 2μC and 3μC placed 10 cm apart in air.

Given: q₁ = 2μC = 2×10⁻⁶ C, q₂ = 3μC = 3×10⁻⁶ C, r = 10 cm = 0.1 m
F = kq₁q₂/r² = (9×10⁹ × 2×10⁻⁶ × 3×10⁻⁶)/(0.1)²
= (54×10⁻³)/0.01 = 5.4 N (repulsive)

2. An electric dipole with dipole moment 4×10⁻⁹ Cm is placed at 30° with a uniform electric field of 5×10⁴ N/C. Calculate the torque.

Given: p = 4×10⁻⁹ Cm, E = 5×10⁴ N/C, θ = 30°
Torque τ = pEsinθ = 4×10⁻⁹ × 5×10⁴ × sin30°
= 20×10⁻⁵ × 0.5 = 10⁻⁴ Nm

3. Three charges of 2μC, -3μC and 4μC are placed at the vertices of an equilateral triangle of side 20 cm. Find the net force on the 4μC charge.

F₁ = k(4μ)(2μ)/(0.2)² = (9×10⁹×8×10⁻¹²)/0.04 = 1.8 N (repulsion along side)
F₂ = k(4μ)(3μ)/(0.2)² = (9×10⁹×12×10⁻¹²)/0.04 = 2.7 N (attraction along other side)
Angle between forces = 60°
F_net = √(F₁² + F₂² + 2F₁F₂cos60°) = √(3.24 + 7.29 + 4.86) ≈ 3.93 N

4. A spherical conductor of radius 12 cm has a charge of 1.6×10⁻⁷ C. Calculate the electric field at 10 cm and 15 cm from its center.

At 10 cm (inside): E = 0 (field inside conductor is zero)
At 15 cm (outside): E = kQ/r² = (9×10⁹ × 1.6×10⁻⁷)/(0.15)²
= 1.44×10³/0.0225 ≈ 6.4×10⁴ N/C radially outward

5. Two point charges of +5μC and -10μC are placed 20 cm apart. Find the point where electric potential is zero.

Let x be distance from +5μC where V=0
k(5μ)/x + k(-10μ)/(20-x) = 0 ⇒ 5/x = 10/(20-x)
5(20-x) = 10x ⇒ 100-5x = 10x ⇒ 100 = 15x ⇒ x = 6.67 cm from +5μC charge

6. A parallel plate capacitor has plates of area 100 cm² separated by 1 mm. Calculate its capacitance if the dielectric medium has εᵣ = 4.

A = 100 cm² = 10⁻² m², d = 1 mm = 10⁻³ m, εᵣ = 4
C = ε₀εᵣA/d = (8.85×10⁻¹² × 4 × 10⁻²)/10⁻³
= 35.4×10⁻¹¹ = 3.54×10⁻¹⁰ F = 354 pF

7. Calculate the energy stored in a 100 pF capacitor when charged to 400 V.

U = ½CV² = ½ × 100×10⁻¹² × (400)²
= 50×10⁻¹² × 160000 = 8×10⁻⁶ J = 8 μJ

Long Answer Questions (5)

1. Derive an expression for electric field intensity at a point on the axial line of an electric dipole.

Derivation:
1. Consider dipole of charges +q and -q separated by 2a
2. At axial point P distance r from center:
E₁ = kq/(r-a)² (from +q), E₂ = kq/(r+a)² (toward -q)
3. Net field E = E₁ - E₂ = kq[1/(r-a)² - 1/(r+a)²]
4. Simplify: E = kq[(r²+2ar+a²)-(r²-2ar+a²)]/[(r-a)(r+a)]²
5. E = kq[4ar]/(r²-a²)²
6. For r >> a: E ≈ 4karq/(r²)² = 2k(2aq)/r³ = 2kp/r³ (since p = 2aq)
7. In vector form: E⃗ = 2kp⃗/r³ (direction same as dipole moment)

2. State and prove Gauss's theorem in electrostatics. Apply it to find electric field due to an infinite plane sheet of charge.

Gauss's Theorem: The total electric flux through any closed surface equals 1/ε₀ times the net charge enclosed.
Proof:
1. Start with Coulomb's law: E = kq/r²
2. Flux through spherical surface: Φ = ∮E⃗·dA⃗ = E×4πr² = (kq/r²)×4πr² = q/ε₀
3. Generalize to any shape by considering solid angles

Application:
1. For infinite sheet with charge density σ, choose cylindrical Gaussian surface
2. Flux through curved surface = 0 (E parallel to surface)
3. Flux through each end = EA
4. Total flux = 2EA = Q_enclosed/ε₀ = σA/ε₀
5. Therefore, E = σ/2ε₀ (constant, independent of distance)

3. Derive the expression for capacitance of a parallel plate capacitor with and without dielectric medium.

Without dielectric:
1. Potential difference V = Ed = (σ/ε₀)d = (Q/Aε₀)d
2. Therefore, Q/V = Aε₀/d ⇒ C = ε₀A/d

With dielectric (K):
1. Field reduces to E = σ/(Kε₀)
2. V = Ed = σd/(Kε₀) = Qd/(Aε₀K)
3. Therefore, C = Q/V = Aε₀K/d = KC₀
4. Where C₀ is capacitance without dielectric

4. Explain the principle, construction and working of a Van de Graaff generator with diagram.

Principle: Electrostatic induction and corona discharge to accumulate very high potential on a metal sphere.
Construction:
1. Large hollow metal sphere on insulated stand
2. Rubber belt moving over two pulleys (lower and upper)
3. Comb-shaped metal electrodes near each pulley
4. Lower comb connected to high voltage source (~10kV)
5. Upper comb connected to inner surface of sphere

Working:
1. Lower comb ionizes air, sprays +ve charges onto belt
2. Belt carries charges upward
3. Upper comb induces -ve charges on inner sphere surface, attracts +ve charges from belt
4. Charges transfer to outer sphere surface via corona discharge
5. Process continues, accumulating high potential (~several million volts)
Note: Diagram would show sphere, belt, pulleys, combs, and charge movement.

5. What is electric potential? Derive an expression for potential at any point due to a point charge.

Electric Potential: Work done per unit charge to bring a test charge from infinity to that point without acceleration.

Derivation for point charge:
1. Potential difference dV = -E⃗·dr⃗
2. For point charge: E = kq/r² radially outward
3. Therefore, dV = -Edr = -kq/r² dr
4. Integrate from ∞ to r: V = -∫(kq/r²)dr = kq/r (evaluated from ∞ to r)
5. V = kq[1/r - 1/∞] = kq/r
6. Absolute potential at distance r: V = (1/4πε₀)q/r

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Chapter 2: Electric Current

Short Answer Questions (10)

1. Define electric current and give its SI unit.

Electric current is the rate of flow of electric charge. I = ΔQ/Δt. Its SI unit is ampere (A), where 1A = 1 coulomb/second.

2. What is drift velocity of electrons? How is it related to current?

Drift velocity is the average velocity of electrons in a conductor under electric field. Current I = nAevₕ where n=electron density, A=cross-section, e=electron charge, vₕ=drift velocity.

3. State Ohm's law. When does it fail?

Ohm's law states V ∝ I or V = IR. It fails for non-ohmic materials (diodes, transistors) and when physical conditions (temperature, etc.) change.

4. Differentiate between emf and terminal voltage.

EMF is the maximum potential difference when no current flows. Terminal voltage is actual voltage across terminals when current flows, given by V = E - Ir (r=internal resistance).

5. What is resistivity? How does it vary with temperature?

Resistivity (ρ) is material's resistance per unit length/area. For metals, ρ increases with temperature (ρₜ = ρ₀[1+αΔT]). For semiconductors, ρ decreases with temperature.

6. Explain Kirchhoff's laws with examples.

Junction Law: ΣI = 0 at any junction (conservation of charge)
Loop Law: ΣV = 0 in any closed loop (conservation of energy)
Example: In parallel resistors, I_total = I₁ + I₂ (junction law). In series, V = V₁ + V₂ (loop law).

7. What is Joule's heating effect? Write its mathematical expression.

Heat produced when current flows through resistor: H = I²Rt (Joules) or H = V²t/R. Power P = I²R = VI = V²/R.

8. Define temperature coefficient of resistance.

It's the fractional change in resistance per degree temperature change: α = (Rₜ - R₀)/(R₀ΔT). Unit: °C⁻¹ or K⁻¹.

9. What is a potentiometer? State its principle.

A potentiometer measures unknown emf by balancing it against potential difference across known length of uniform wire. Principle: V ∝ L when constant current flows.

10. Differentiate between series and parallel combinations of resistors.

Series: Same current, V_total = V₁ + V₂, R_eq = R₁ + R₂
Parallel: Same voltage, I_total = I₁ + I₂, 1/R_eq = 1/R₁ + 1/R₂

Numerical Problems (7)

1. A current of 0.5 A flows through a wire for 6 minutes. Calculate the charge that passes through any cross-section.

I = 0.5A, t = 6 min = 360s
Q = It = 0.5 × 360 = 180 C

2. A wire of resistance 5Ω is stretched to double its length. Calculate its new resistance.

Volume remains same: A₁L₁ = A₂L₂ ⇒ A₂ = A₁/2 (since L₂ = 2L₁)
R = ρL/A ⇒ R₂ = ρ(2L₁)/(A₁/2) = 4(ρL₁/A₁) = 4R₁ = 20Ω

3. Calculate the equivalent resistance between points A and B in a network of three 6Ω resistors connected in triangle.

Delta connection can be converted to star: each star arm = (6×6)/(6+6+6) = 2Ω
R_eq = 2 + (2+6)∥2 = 2 + (8×2)/(8+2) = 2 + 1.6 = 3.6Ω

4. A cell of emf 2V and internal resistance 0.5Ω is connected to a 3.5Ω resistor. Calculate the terminal voltage.

I = E/(R+r) = 2/(3.5+0.5) = 0.5A
Terminal voltage V = E - Ir = 2 - (0.5×0.5) = 1.75V

5. An electric heater rated 1000W, 220V is connected to 110V supply. Calculate the power consumed.

Resistance R = V²/P = (220)²/1000 = 48.4Ω
At 110V: P = V²/R = (110)²/48.4 = 250W

6. In a meter bridge, the balance point is found at 40 cm from end A when resistance in left gap is 2Ω. Calculate the unknown resistance.

For meter bridge: R/S = l/(100-l)
2/X = 40/60 ⇒ X = 3Ω

7. A potentiometer wire of length 4m has resistance 8Ω. Find the emf of a cell which balances at 250 cm when a standard cell of 1.2V balances at 200 cm.

E₁/E₂ = l₁/l₂ ⇒ E/1.2 = 250/200
E = 1.2 × (250/200) = 1.5V

Long Answer Questions (5)

1. Derive the expression for drift velocity of electrons and establish relation between current and drift velocity.

Derivation:
1. In time Δt, electrons travel distance Δx = vₕΔt
2. Number of electrons in volume AΔx = nAΔx = nAvₕΔt
3. Charge ΔQ = (nAvₕΔt)e
4. Current I = ΔQ/Δt = nAevₕ
5. Therefore vₕ = I/(nAe)
Where n=electron density, A=area, e=electron charge

2. Derive the condition for balanced Wheatstone bridge using Kirchhoff's laws.

Derivation:
1. Apply junction law at point C: I₁ = I₃ and I₂ = I₄
2. Apply loop law to ABDA: I₁P - I₂R = 0 ⇒ P/R = I₂/I₁
3. Apply loop law to BCDB: I₃Q - I₄S = 0 ⇒ Q/S = I₄/I₃
4. Since I₁ = I₃ and I₂ = I₄, we get P/R = Q/S or P/Q = R/S
This is the balanced condition for Wheatstone bridge

3. Explain the principle, construction and working of a potentiometer. How is it superior to voltmeter?

Principle: Potential difference across any length of uniform wire is proportional to its length when constant current flows.
Construction:
1. Long uniform wire (usually 4-10m) of high resistivity (constantan/manganin)
2. Wooden board with scale
3. Battery, rheostat, key for driving current
4. Jockey for making contact
Working:
1. Standard cell balances at length l₁
2. Unknown cell balances at l₂ ⇒ E₂ = (l₂/l₁)E₁
Advantages over voltmeter:
1. No current drawn during measurement (more accurate)
2. Can measure small potential differences precisely

4. Derive expressions for equivalent resistance in series and parallel combinations of resistors.

Series Combination:
1. Same current I flows through all resistors
2. Total V = V₁ + V₂ = IR₁ + IR₂ = I(R₁ + R₂)
3. Therefore R_eq = V/I = R₁ + R₂

Parallel Combination:
1. Same voltage V across all resistors
2. Total I = I₁ + I₂ = V/R₁ + V/R₂ = V(1/R₁ + 1/R₂)
3. Therefore 1/R_eq = I/V = 1/R₁ + 1/R₂

5. What is a meter bridge? Explain how it can be used to determine unknown resistance.

Meter Bridge: A Wheatstone bridge implementation using 1m long uniform wire.
Construction:
1. Wire AC of 1m length with scale
2. Known resistance R in left gap
3. Unknown resistance S in right gap
4. Galvanometer and jockey
Working:
1. Find balance point D where galvanometer shows null deflection
2. Measure lengths AD = l and DC = (100-l) cm
3. Apply Wheatstone bridge condition: R/S = l/(100-l)
4. Therefore S = R(100-l)/l
Precautions:
1. Ensure tight connections
2. Take multiple readings
3. Use high resistance in series with galvanometer initially

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Chapter 3: Magnetism

Short Answer Questions (10)

1. Define magnetic dipole moment. Give its SI unit.

Magnetic dipole moment (m) is the product of pole strength (m) and length of magnet (2l). m = m × 2l. SI unit: Am² or JT⁻¹.

2. State Biot-Savart law in vector form.

dB⃗ = (μ₀/4π) (Idl⃗×r̂)/r² where dB⃗ is magnetic field, I is current, dl⃗ is length element, r̂ is unit vector from dl to point, and r is distance.

3. What is Ampere's circuital law?

∮B⃗·dl⃗ = μ₀I_enclosed, meaning line integral of B around any closed path equals μ₀ times current enclosed by the path.

4. Differentiate between diamagnetic, paramagnetic and ferromagnetic materials.

Diamagnetic: Weakly repelled (μᵣ < 1)
Paramagnetic: Weakly attracted (μᵣ > 1)
Ferromagnetic: Strongly attracted (μᵣ ≫ 1), retains magnetism

5. What is the force between two parallel current carrying conductors?

F/L = (μ₀I₁I₂)/(2πd) per unit length. Attractive for same direction currents, repulsive for opposite directions.

6. Define magnetic susceptibility and relative permeability.

Magnetic susceptibility (χₘ): Ratio of magnetization to applied field
Relative permeability (μᵣ): Ratio of material's permeability to μ₀ (μᵣ = 1 + χₘ)

7. What is Lorentz force? Write its expression.

Force on charged particle moving in EM field: F⃗ = q(E⃗ + v⃗×B⃗). Magnetic part: F⃗ = q(v⃗×B⃗).

8. Explain the principle of a moving coil galvanometer.

Torque τ = nIABsinθ acts on current-carrying coil in magnetic field, causing proportional deflection θ ∝ I (for small angles).

9. What is cyclotron frequency? Give its expression.

Frequency of charged particle in magnetic field: f = qB/(2πm). Independent of velocity (for v ≪ c).

10. State Curie's law for paramagnetic substances.

χₘ ∝ 1/T or χₘ = C/T where C is Curie constant, T is absolute temperature.

Numerical Problems (7)

1. A current of 10A flows through a circular coil of radius 10 cm having 100 turns. Calculate the magnetic field at its center.

B = μ₀NI/(2r) = (4π×10⁻⁷×100×10)/(2×0.1)
= 2π×10⁻³ ≈ 6.28×10⁻³ T (or 6.28 mT)

2. A proton moves with a speed of 5×10⁶ m/s perpendicular to a magnetic field of 0.2T. Calculate the force on the proton.

F = qvBsin90° = (1.6×10⁻¹⁹)(5×10⁶)(0.2)
= 1.6×10⁻¹³ N (direction given by right-hand rule)

3. Two long parallel wires 10 cm apart carry currents of 5A and 10A in same direction. Calculate the force per unit length between them.

F/L = (μ₀I₁I₂)/(2πd) = (4π×10⁻⁷×5×10)/(2π×0.1)
= 10⁻⁴ N/m (attractive since currents are parallel)

4. A bar magnet of magnetic moment 1.5 JT⁻¹ is placed at 30° with a uniform magnetic field of 0.22 T. Calculate the torque.

τ = mBsinθ = 1.5 × 0.22 × sin30°
= 1.5 × 0.22 × 0.5 = 0.165 Nm

5. A solenoid of length 50 cm has 500 turns. If it carries current 5A, calculate the magnetic field inside it.

B = μ₀nI = μ₀(N/L)I
= (4π×10⁻⁷)(500/0.5)(5) = 2π×10⁻³ ≈ 6.28×10⁻³ T

6. An electron moves in a circular path of radius 2 cm in a magnetic field of 2×10⁻³ T. Calculate its speed.

mv²/r = qvB ⇒ v = qBr/m
= (1.6×10⁻¹⁹×2×10⁻³×0.02)/(9.1×10⁻³¹)
≈ 7.0×10⁶ m/s

7. A galvanometer has resistance 50Ω and gives full scale deflection for 5mA current. How will you convert it into a voltmeter of 0-10V range?

V = I_g(R_g + R) ⇒ 10 = 0.005(50 + R)
R = (10/0.005) - 50 = 2000 - 50 = 1950Ω
Connect 1950Ω series resistance

Long Answer Questions (5)

1. Derive an expression for magnetic field due to a current carrying circular loop at its center.

Derivation:
1. Biot-Savart law: dB = (μ₀/4π)(Idlsinθ)/r²
2. For circular loop: θ = 90°, sinθ = 1, r = R (radius)
3. dB = (μ₀/4π)(Idl)/R²
4. All dl components add up along axis: B = ∮dBsin90°
5. B = (μ₀/4π)(I/R²)∮dl = (μ₀/4π)(I/R²)(2πR)
6. Therefore B = μ₀I/(2R) at center
For N turns: B = μ₀NI/(2R)

2. Derive the expression for force on a current carrying conductor in a magnetic field.

Derivation:
1. Force on charge q: F = q(v⃗×B⃗)
2. For n charges per unit volume: dF = nAdl(qv⃗×B⃗)
3. Current I = nqvA ⇒ nqv = I/A
4. dF = (I/A)Adl(dl̂×B⃗) = I(dl⃗×B⃗)
5. For finite length: F = I∫(dl⃗×B⃗)
6. For uniform B and straight wire: F = IlBsinθ

3. What is a cyclotron? Explain its principle, construction and working with diagram.

Principle: Charged particles accelerate in electric field while magnetic field makes them follow circular path.
Construction:
1. Two D-shaped hollow metal dees
2. Strong electromagnet
3. High frequency oscillator
4. Vacuum chamber
Working:
1. Particles injected at center
2. Alternating electric field accelerates particles each half-cycle
3. Magnetic field causes circular motion with radius r = mv/qB
4. Kinetic energy increases with each crossing
Note: Diagram would show dees, magnetic field, particle path, and alternating electric field.

4. Derive an expression for torque on a current carrying loop in a uniform magnetic field.

Derivation:
1. Consider rectangular loop of sides a and b
2. Forces on sides || to B cancel out
3. Forces on ⊥ sides: F = IbB (each)
4. Torque τ = F × (a/2)sinθ × 2 = IbB × a sinθ
5. Since area A = ab: τ = IABsinθ
6. For N turns: τ = NIABsinθ
7. In vector form: τ⃗ = m⃗×B⃗ where m⃗ = NIA⃗

5. Explain the theory, construction and working of a moving coil galvanometer. Derive the expression for current sensitivity.

Theory: Torque on current-carrying coil in magnetic field causes proportional deflection.
Construction:
1. Rectangular coil suspended between pole pieces
2. Radial magnetic field
3. Phosphor bronze strip provides restoring torque
4. Soft iron core increases field strength
Working:
1. Current I produces torque τ = NIAB
2. Restoring torque τᵣ = kθ (k = torsion constant)
3. At equilibrium: NIAB = kθ ⇒ θ = (NAB/k)I
Current sensitivity: Sᵢ = θ/I = NAB/k (radians per ampere)

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Chapter 4: Electromagnetic Induction and Alternating Current

Short Answer Questions (10)

1. State Faraday's laws of electromagnetic induction.

First Law: When magnetic flux through a circuit changes, an emf is induced
Second Law: Magnitude of induced emf is proportional to rate of change of flux (ε = -dΦ/dt)

2. What is Lenz's law? How does it follow conservation of energy?

Lenz's law states that induced current opposes the change causing it. This conserves energy because work must be done against the opposing force to change the flux.

3. Define self inductance and mutual inductance.

Self inductance (L): Property of coil to oppose change in its own current (ε = -L di/dt)
Mutual inductance (M): Property where change in current in one coil induces emf in nearby coil (ε₂ = -M di₁/dt)

4. What is eddy current? Give one useful application.

Eddy currents are circulating currents induced in bulk metals when exposed to changing magnetic fields. Application: Induction furnace for metal heating.

5. Differentiate between step-up and step-down transformers.

Step-up: Increases voltage (Nₛ > Nₚ, Vₛ > Vₚ)
Step-down: Decreases voltage (Nₛ < Nₚ, Vₛ < Vₚ)
Both follow Vₛ/Vₚ = Nₛ/Nₚ = Iₚ/Iₛ (for ideal transformer)

6. Define RMS value of alternating current.

RMS value is the equivalent DC current that would produce same heating effect. For sinusoidal AC, Iᵣₘₛ = I₀/√2, where I₀ is peak current.

7. What is power factor? What is its value for pure resistive circuit?

Power factor = cosφ = R/Z (ratio of real power to apparent power). For pure resistive circuit, φ=0 ⇒ cosφ=1 (maximum).

8. State the principle of AC generator.

AC generator works on electromagnetic induction - when coil rotates in magnetic field, changing flux induces alternating emf ε = NBAωsinωt.

9. What are wattless currents?

Current component (I₀sinφ) that is 90° out of phase with voltage, contributing no power (P = VIcosφ). Occurs in purely inductive/capacitive circuits.

10. Why is iron core laminated in transformers?

Laminations reduce eddy currents by increasing resistance, thereby minimizing energy losses due to heating (P = V²/R).

Numerical Problems (7)

1. A coil of 100 turns is moved in 0.5s between poles of a magnet where flux changes from 8×10⁻⁴ Wb to 3×10⁻⁴ Wb. Calculate the induced emf.

ε = -NΔΦ/Δt = -100×(3×10⁻⁴ - 8×10⁻⁴)/0.5
= -100×(-5×10⁻⁴)/0.5 = 0.1 V

2. An aircraft with 20m wingspan flies at 720 km/h perpendicular to Earth's magnetic field (5×10⁻⁵ T). Calculate the potential difference across wings.

v = 720 km/h = 200 m/s
ε = Blv = 5×10⁻⁵ × 20 × 200 = 0.2 V

3. A solenoid of length 50 cm and 500 turns has self inductance 0.5H. Calculate the area of cross-section.

L = μ₀N²A/l ⇒ A = Ll/(μ₀N²)
= (0.5×0.5)/(4π×10⁻⁷×500²) ≈ 0.0008 m² = 8 cm²

4. An AC voltage V = 220 sin(100πt) is applied across a 100Ω resistor. Calculate RMS current and power dissipated.

V₀ = 220V ⇒ Vᵣₘₛ = 220/√2 ≈ 155.6 V
Iᵣₘₛ = Vᵣₘₛ/R = 155.6/100 = 1.556 A
P = VᵣₘₛIᵣₘₛ = 155.6 × 1.556 ≈ 242 W

5. A transformer has 200 primary turns and 50 secondary turns. If primary voltage is 220V, calculate secondary voltage.

Vₛ/Vₚ = Nₛ/Nₚ ⇒ Vₛ = (Nₛ/Nₚ)Vₚ
= (50/200)×220 = 55 V

6. A series LCR circuit has R=10Ω, L=0.1H and C=10μF. Calculate the resonant frequency.

ωᵣ = 1/√(LC) = 1/√(0.1×10×10⁻⁶) = 1000 rad/s
fᵣ = ωᵣ/2π ≈ 159.2 Hz

7. An inductor of 0.5H is connected to 220V, 50Hz AC supply. Calculate the inductive reactance and RMS current.

X_L = 2πfL = 2π×50×0.5 ≈ 157 Ω
Iᵣₘₛ = Vᵣₘₛ/X_L = 220/157 ≈ 1.4 A

Long Answer Questions (5)

1. State and explain Faraday's laws of electromagnetic induction. Derive expression for induced emf.

Faraday's First Law: When magnetic flux through circuit changes, emf is induced
Faraday's Second Law: ε = -dΦ/dt

Derivation:
1. Work done to move conductor: W = F·d = (IlB) × vΔt
2. emf ε = W/q = (IlBvΔt)/(IΔt) = Blv
3. Rate of change of area dA/dt = lv
4. Change in flux dΦ = B·dA ⇒ dΦ/dt = Blv
5. Therefore ε = dΦ/dt (magnitude)
6. Lenz's law gives negative sign: ε = -dΦ/dt

2. Derive an expression for self inductance of a long solenoid.

Derivation:
1. Magnetic field inside solenoid: B = μ₀nI (n = N/l)
2. Flux through one turn: Φ₁ = BA = μ₀nIA
3. Total flux linkage: Φ = NΦ₁ = μ₀nNIA
4. But n = N/l ⇒ Φ = μ₀(N²/l)IA
5. Self inductance L = Φ/I = μ₀N²A/l
6. For solenoid with core: L = μ₀μᵣN²A/l

3. What is a transformer? Derive the expression for transformer ratio in terms of number of turns and currents.

Transformer: Device that changes AC voltage using mutual induction

Derivation:
1. Primary flux Φₚ = Secondary flux Φₛ (ideal transformer)
2. εₚ = -NₚdΦ/dt, εₛ = -NₛdΦ/dt
3. Therefore εₛ/εₚ = Nₛ/Nₚ
4. For ideal transformer: Input power = Output power ⇒ VₚIₚ = VₛIₛ
5. Thus Vₛ/Vₚ = Iₚ/Iₛ = Nₛ/Nₚ

4. Derive the expression for impedance and phase angle in series LCR circuit. Draw phasor diagram.

Derivation:
1. V_R = IR, V_L = IX_L, V_C = IX_C
2. Net voltage V = √[V_R² + (V_L - V_C)²]
3. Therefore Z = V/I = √[R² + (X_L - X_C)²]
4. Phase angle tanφ = (V_L - V_C)/V_R = (X_L - X_C)/R

Phasor Diagram:
- V_R along x-axis
- V_L 90° upwards
- V_C 90° downwards
- Resultant V at angle φ

5. Explain the principle, construction and working of an AC generator with diagram.

Principle: Electromagnetic induction - changing flux induces emf

Construction:
1. Armature coil (N turns, area A)
2. Strong field magnet
3. Slip rings and brushes
4. Rotating shaft

Working:
1. Coil rotates with angular velocity ω in field B
2. Flux Φ = NBAcosθ = NBAcosωt
3. Induced emf ε = -dΦ/dt = NBAωsinωt
4. ε₀ = NBAω ⇒ ε = ε₀sinωt (alternating emf)

Note: Diagram would show coil, magnetic field, slip rings, brushes, and output waveform.

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Chapter 5: Electromagnetic Waves

Short Answer Questions (10)

1. What are electromagnetic waves? Name the scientist who first predicted their existence.

EM waves are synchronized oscillations of electric and magnetic fields that propagate through space. James Clerk Maxwell first predicted their existence theoretically.

2. Write the basic source of electromagnetic waves.

Accelerated charged particles produce EM waves. Common sources include oscillating charges in antennas or atomic transitions.

3. State the four Maxwell's equations in integral form.

1. ∮E⃗·dA⃗ = Q/ε₀ (Gauss's law for electricity)
2. ∮B⃗·dA⃗ = 0 (Gauss's law for magnetism)
3. ∮E⃗·dl⃗ = -dΦ_B/dt (Faraday's law)
4. ∮B⃗·dl⃗ = μ₀I + μ₀ε₀dΦ_E/dt (Ampere-Maxwell law)

4. Write the expression for speed of electromagnetic waves in vacuum.

c = 1/√(μ₀ε₀) ≈ 3×10⁸ m/s, where μ₀ is permeability and ε₀ is permittivity of free space.

5. What is the ratio of magnitudes of electric and magnetic fields in an EM wave?

E₀/B₀ = c (speed of light in vacuum). In SI units, E = cB.

6. Name the physical quantity which remains same for X-rays and radio waves in vacuum.

Speed (c ≈ 3×10⁸ m/s). All EM waves travel at same speed in vacuum, differing only in frequency/wavelength.

7. What is the relation between energy density of electric and magnetic fields in EM wave?

Energy densities are equal: u_E = ½ε₀E² = u_B = B²/(2μ₀). Total energy density u = ε₀E² = B²/μ₀.

8. Why are microwaves considered suitable for radar systems?

Microwaves (1mm-1m) are suitable because they: (1) pass through atmosphere with little scattering, (2) provide good resolution, (3) can be focused into narrow beams.

9. What is the range of visible spectrum in terms of wavelength?

Approximately 400 nm (violet) to 700 nm (red). Frequency range: 7.5×10¹⁴ Hz to 4.3×10¹⁴ Hz.

10. Why can't we hear radio waves?

Human ear detects sound waves (mechanical, 20Hz-20kHz), while radio waves are EM waves (10⁴-10¹¹Hz) that require electronic conversion to audible signals.

Numerical Problems (7)

1. Calculate the wavelength of EM wave whose frequency is 900 kHz.

λ = c/f = (3×10⁸)/(900×10³) ≈ 333.3 m (radio wave)

2. The magnetic field amplitude of EM wave is 2×10⁻⁷ T. Calculate electric field amplitude.

E₀ = cB₀ = 3×10⁸ × 2×10⁻⁷ = 60 V/m

3. An EM wave has electric field E = 3.1×10²sin(1.8×10⁶x - 5.4×10¹⁴t) V/m. Find its wavelength.

Comparing with E = E₀sin(kx-ωt):
k = 2π/λ = 1.8×10⁶ ⇒ λ = 2π/1.8×10⁶ ≈ 3.49×10⁻⁶ m (UV)

4. Calculate energy density of EM wave having E = 50 V/m.

u = ε₀E² = 8.85×10⁻¹² × (50)² ≈ 2.21×10⁻⁸ J/m³

5. A radio station broadcasts at 100 MHz. How long does signal take to reach receiver 30 km away?

t = d/c = 30×10³/(3×10⁸) = 10⁻⁴ s = 0.1 ms

6. The amplitude of magnetic field in EM wave is 6 μT. Calculate intensity of wave.

I = (cB₀²)/(2μ₀) = (3×10⁸×(6×10⁻⁶)²)/(2×4π×10⁻⁷) ≈ 0.43 W/m²

7. An EM wave has wavelength 500 nm. Find its angular frequency.

ω = 2πc/λ = 2π×3×10⁸/(500×10⁻⁹) ≈ 3.77×10¹⁵ rad/s (visible light)

Long Answer Questions (5)

1. Describe Hertz's experiment to demonstrate existence of electromagnetic waves.

Apparatus:
1. Induction coil connected to metal plates with spark gap (transmitter)
2. Loop antenna with smaller spark gap (receiver)

Procedure:
1. High voltage across transmitter gap creates sparks
2. Oscillating charges produce EM waves (predicted by Maxwell)
3. Distant receiver loop shows tiny sparks synchronously

Observations:
1. Sparks only when receiver loop properly oriented
2. Standing wave pattern when reflector used
3. Measured wavelength matched Maxwell's predictions

Significance: First experimental confirmation of EM waves (1887)

2. Derive the expression for speed of electromagnetic waves from Maxwell's equations.

Derivation:
1. Start with Maxwell's equations in vacuum (no charges/currents)
2. Take curl of Faraday's law: ∇×(∇×E) = -∂(∇×B)/∂t
3. Use vector identity and Ampere-Maxwell law:
∇(∇·E) - ∇²E = -μ₀ε₀∂²E/∂t²
4. In vacuum ∇·E = 0 ⇒ ∇²E = μ₀ε₀∂²E/∂t² (wave equation)
5. Comparing with standard wave equation: speed c = 1/√(μ₀ε₀)
6. Calculated value matches speed of light ⇒ light is EM wave

3. Describe electromagnetic spectrum with frequency range and applications of each band.

Type	Wavelength	Frequency	Applications
Radio	>1m	<3×10⁸Hz	Broadcasting, communication
Microwave	1mm-1m	3×10⁸-3×10¹¹Hz	Radar, satellite comm.
Infrared	700nm-1mm	3×10¹¹-4×10¹⁴Hz	Thermal imaging, remote controls
Visible	400-700nm	4-7.5×10¹⁴Hz	Vision, photography
UV	10-400nm	7.5×10¹⁴-3×10¹⁶Hz	Sterilization, fluorescence
X-rays	0.01-10nm	3×10¹⁶-3×10¹⁹Hz	Medical imaging, crystallography
Gamma	<0.01nm	>3×10¹⁹Hz	Cancer treatment, nuclear physics

4. Show that the average value of intensity of EM wave is given by I = ½cε₀E₀².

Derivation:
1. Instantaneous energy density: u = ε₀E² = ε₀E₀²sin²(kx-ωt)
2. Average over time: ⟨sin²θ⟩ = ½ ⇒ u_avg = ½ε₀E₀²
3. Intensity = energy crossing unit area per unit time = cu_avg
4. Therefore I = ½cε₀E₀²

Alternative using Poynting vector:
1. S⃗ = E⃗×H⃗ = (1/μ₀)E⃗×B⃗
2. For EM wave: |S| = EB/μ₀ = E²/(μ₀c)
3. Using c = 1/√(μ₀ε₀) ⇒ I_avg = ½cε₀E₀²

5. Explain the properties of electromagnetic waves.

Properties:
1. Transverse waves: E⃗ and B⃗ oscillate perpendicular to propagation direction
2. Orthogonal oscillations: E⃗ ⊥ B⃗ ⊥ propagation direction
3. Speed in vacuum: Constant c ≈ 3×10⁸ m/s (independent of frequency)
4. Energy transport: Carries energy E = hν (quantized as photons)
5. Momentum: Exerts radiation pressure p = I/c
6. Polarization: Can be linearly, circularly or elliptically polarized
7. Reflection/refraction: Follows laws of reflection and refraction
8. Interference/diffraction: Exhibits wave phenomena
9. No medium required: Propagates through vacuum
10. Spectrum: Continuous range of frequencies/wavelengths

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