SOLUTIONS
2026 BOARD EXAM
CBSE Class 12 PCM Physics Board Paper 2026 — Set 2 Solutions
2026
Physics
Class 12 PCM
CBSE EXAMINATION PAPER-2026
PHYSICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 38 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 2 are case based questions
- Section B – questions number 3 to 18 are multiple choice questions
- Section C – questions number 19 to 24 are very short answer
- Section D – questions number 25 to 32 are short answer
- Section E – questions number 33 to 38 are long answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
(1) Silicon is doped with which of the following to obtain p-type semiconductor? (A) Phosphorus (B) Arsenic (C) Boron (D) Antimony
(2) A semiconductor has an electron concentration of 5 × 10²² m⁻³. The concentration of holes is (given nᵢ = 1.5 × 10¹⁶ m⁻³) (A) 5 × 10²² m⁻³ (B) 1.5 × 10⁶ m⁻³ (C) 9 × 10⁸ m⁻³ (D) 4.5 × 10⁹ m⁻³
(3) During forward biasing of a p-n junction diode, the (A) current is mainly due to drifting of majority carriers (B) current is mainly due to drifting of minority carriers (C) diffusion and drift currents are equal (D) current is of the order of 1 A
(4) The threshold voltage for silicon diode is about (A) 0.2 V (B) 0.5 V (C) 0.7 V (D) 1.5 V
This is the minimum forward voltage at which the diode starts to conduct.
This value is based on the energy barrier created at the p-n junction, as explained in the case paragraph about barrier electric field and diffusion current.
(5) When we dope Ge with a pentavalent element, four of its electrons bond with four germanium neighbours but fifth electron remains weakly bound. The ionisation energy for this electron is about (A) 0.01 eV (B) 0.05 eV (C) 0.1 eV (D) 0.15 eV
(1) The value of d is (A) √D(D − 4f) (B) √D(D − 2f) (C) 2√(Df) (D) √D(D − f)
Correct answer is option (A)
According to the lens displacement method, the focal length is given by the formula: f = (D^2 - d^2) / 4D. Rearranging this equation to solve for d gives d = √[D(D - 4f)].
(2) Compared to the size of the object, the images formed in the two positions are respectively (A) reduced, enlarged (B) reduced, reduced (C) enlarged, enlarged (D) enlarged, reduced
Correct answer is option (d)
When the lens is closer to the object, the image distance is greater than the object distance, creating an enlarged image. As the lens moves closer to the screen, the object distance becomes greater than the image distance, creating a reduced image.
(3) If the distance between object and screen is 80.00 cm and the lens forms sharp images at two positions separated by 20.00 cm, the focal length of convex lens is (A) 15.50 cm (B) 18.75 cm (C) 20.50 cm (D) 22.75 cm
(4) Consider a convex lens of focal length 15 cm. For which of the following values of object-screen distance, two positions of the lens can be found to obtain sharp image on the screen? (A) 45 cm (B) 50 cm (C) 55 cm (D) 65 cm
Correct answer is Option (D)
For two real distinct images to form on the screen in the displacement method, the distance between the object and the screen must be strictly greater than four times the focal length (D > 4f).
4f = 4 * 15 cm = 60 cm
only 65 cm is greater than 60 cm.
(5) A thin convex lens of focal length 10 cm and another thin lens of focal length ‘f’ are placed coaxially in contact. If the power of their combination is 10/ 3 D, the value of ‘f’ is (A)–15 cm (B) –10 cm (C) –20 cm (D) –30 cm
Section B
Three point charges 2q, –2q and q are kept at the vertices of an equilateral triangle of side l. The potential energy of the system is
Two metal spheres of radii r₁ and r₂ (r₁ > r₂) having charges q₁ and q₂ respectively kept in air, are brought in contact. Which of the following statements is not correct?
The maximum kinetic energy of the electrons emitted from a photosensitive surface depends on
In Bohr model of hydrogen, for large values of n, the distance between consecutive orbits is proportional to
A straight long wire lying along y-axis, carries a current of 1 A along –y direction. The magnetic field due to the conductor at a point (50 cm, 0, 0) will point along.
Which of the following materials has positive and small value of magnetic susceptibility?
A galvanometer of resistance 27 Ω is converted into an ammeter of range (0 – 10 mA) using a resistance of 3 Ω. The galvanometer will show full scale deflection for a current of about-
The magnetic flux ϕ (in Wb) linked with a coil is related to time t (in s) as ϕ = 5 At² + Bt – 2C The SI units of A and B are respectively
The figure shows the variation of capacitive reactance (X꜀) of two ideal capacitors of capacitances C₁ & C₂ with the reciprocal of angular frequency (1/ω) of ac source. The value of C₁/C₂ is
An electromagnetic wave is propagating along x-axis. At any instant, the phase difference (in radian) between the electric field (𝐄⃗) and the magnetic field (𝐁⃗) associated with the wave is
In Bohr model of hydrogen atom, the electron makes a transition from n = 5 to n = 1 state. As a result, a photon of wavelength λ is emitted in the process. The wavelength of the photon emitted when an electron makes a transition from energy level n = 5 to n = 2 will be
Assertion (A): On increasing the intensity of incident light of frequency (ν > ν₀) on a photosensitive surface, the photocurrent increases. Reason (R): The stopping potential for a photosensitive surface increases with increase of frequency (ν > ν₀) of incident light.
Assertion (A): On forward biasing a p-n junction diode, the height of the barrier potential increases. Reason (R): In forward biasing of a p-n junction diode, the direction of the applied voltage is in the same direction as the built-in potential.
Assertion (A): Two electric heaters of power P₁ and P₂ (P₁ > P₂) are joined in series across a dc source of voltage V. The power consumed by the combination will be less than that consumed by P₁ when connected across the same source. Reason (R): The power consumed by an electric device when connected to a dc source of voltage V is proportional to its resistance.
Section C
Write two points of difference between intrinsic and extrinsic semiconductors.
Find ratio (λₐ/λₚ) of the de Broglie wavelength λₐ and λₚ associated respectively with an alpha particle and a proton,
(i) if they are moving with the same kinetic energy.
(ii) just after they are accelerated through the same potential difference.
We have v = sqrt(2K/m). The de Broglie wavelength λ = h/mv = h/(m*sqrt(2K/m)) = h/sqrt(2Km). So, λ is inversely proportional to sqrt(m).
Therefore, (λₐ/λₚ) = sqrt(mₚ/mₐ)
Since mₐ = 4 mₚ (alpha particle has 2 protons and 2 neutrons),
(λₐ/λₚ) = sqrt(mₚ/4 mₚ) = 1/2
(ii) When accelerated through the same potential difference V, kinetic energy gained K = qV
For alpha particle, charge qₐ = 2e; For proton, qₚ = e
Then Kₐ = 2eV, Kₚ = eV
Velocity v = sqrt(2K/m). So,
λ = h/(m v) = h/(m sqrt(2K/m)) = h/sqrt(2mK)
Therefore, (λₐ/λₚ) = sqrt(mₚ qₚ / mₐ qₐ) = sqrt(mₚ e / 4 mₚ * 2 e) = sqrt(1 / 8) = 1/(2 sqrt(2))
A ray of light in air is incident at angle ∠i on a face of an equilateral glass prism and is refracted through the prism. As ∠i is varied, it is observed that the ray undergoes minimum deviation, when the ∠i is three-fourth of the angle of the prism. Calculate the speed of light in the prism.
At minimum deviation: Angle of incidence i = angle of emergence e, and angles of refraction r1 = r2 = r.
Using prism angle relation, r1 + r2 = A, so 2r = 60 => r = 30deg.
Using Snell's law at first face, n = sin i / sin r = sin 45 / sin 30 = (0.7071) / (0.5) = 1.414.
Refractive index, n = speed of light in air / speed of light in prism => v = c / n = 3 * 108 m/s / 1.414 = 2.12 * 108 m/s.
Answer: Speed of light in the prism is 2.12 * 108 m/s.
A heating element using nichrome is connected to a 220 V supply. Initially it draws a current of 2.9 A. After some time, the current attains a steady value of 2.5 A. Find the steady temperature of the heating element if the room temperature is 27 °C. The temperature coefficient of resistance of nichrome is 1.7 × 10⁻⁴ °C⁻¹.
Voltage, V = 220 V
Initial current, I1 = 2.9 A
Final current, I2 = 2.5 A
Room temperature, T1 = 27 degC
Temperature coefficient of resistance, alpha = 1.7 * 10^-4 degC^-1
Step 1: Calculate initial resistance, R1 = V/I1 = 220/2.9 = 75.86 ohms
Step 2: Calculate final resistance, R2 = V/I2 = 220/2.5 = 88 ohms
Step 3: Use formula for resistance change with temperature:
R2 = R1 * [1 + alpha * (T2 - T1)]
=> 88 = 75.86 * [1 + 1.7 * 10^-4 * (T2 - 27)]
=> (88/75.86) - 1 = 1.7 * 10^-4 * (T2 - 27)
=> 0.16 = 1.7 * 10^-4 * (T2 - 27)
=> T2 - 27 = 0.16 / (1.7 * 10^-4) = 941.18 degC
=> T2 = 27 + 941.18 = 968.18 degC
Therefore, the steady temperature of the heating element is approximately 968 degC.
A 5 cm long pencil is placed along the principal axis of a concave mirror of focal length 20 cm such that its nearest end is at a distance of 25 cm from the mirror. Calculate the length of the image of the pencil.
Step 1: Calculate image distance v for point A using mirror formula 1/f = 1/v + 1/u.
=> 1/v = 1/f - 1/u = 1/(-20) - 1/(-25) = -1/20 + 1/25 = (-5 + 4)/100 = -1/100.
=> v = -100 cm.
Step 2: Now object distance for point B = u + length of pencil = -25 - 5 = -30 cm.
Calculate image distance v' for point B: 1/f = 1/v' + 1/u'.
=> 1/v' = 1/f - 1/u' = 1/(-20) - 1/(-30) = -1/20 + 1/30 = (-3 + 2)/60 = -1/60.
=> v' = -60 cm.
Step 3: Length of image = |v'| - |v| = 100 cm - 60 cm = 40 cm.
Therefore, the length of the image of the pencil is 40 cm.
In a Young's double-slit experiment, a beam of light consisting of two wavelengths 500 nm and 600 nm is used. The interference fringes are observed at a screen placed 1.8 m away from the plane of slits (slit separation 0.3 mm). Calculate the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.
Given: lambda1 = 500 nm = 500 * 10^-9 m, lambda2 = 600 nm = 600 * 10^-9 m, D = 1.8 m, d = 0.3 mm = 0.3 * 10^-3 m.
The bright fringes of both wavelengths coincide where their fringe positions are equal:
(m * lambda1 * D) / d = (n * lambda2 * D) / d => m * lambda1 = n * lambda2
=> m / n = lambda2 / lambda1 = 600 / 500 = 6 / 5
So, the smallest integers m and n satisfying this are m=6, n=5.
Position of coincident bright fringes: x = (m * lambda1 * D) / d = (6 * 500 * 10^-9 * 1.8) / (0.3 * 10^-3) = 0.018 m = 1.8 cm.
Therefore, the least distance from the central maximum where bright fringes of both wavelengths coincide is 1.8 cm.
Section D
(a) Consider the following nuclides:
¹²₆C, ¹⁹⁸₈₀Hg, ¹⁴₆C, ¹⁹⁷₇₉Au
Group them into isotopes and isotones.
(b) How does the size of a nucleus depend on its mass number A? Hence prove that the density of nucleus is a constant, independent of A, for all nuclei.
(b) The size or radius R of a nucleus depends on its mass number A by the relation R = R₀ * A1/3, where R₀ is a constant approximately 1.2 fm (1.2 * 10-15 m). Since volume V of the nucleus is proportional to R3, we get V proportional to A. The mass of the nucleus is proportional to A as it equals A * mass of a nucleon. Therefore, density = mass/volume is proportional to A/A = constant. Hence, nuclear density is constant and independent of A for all nuclei. This means all nuclei have approximately the same density, which is very high compared to ordinary matter.
(a) An electric field 𝐄⃗ = E₀ î exists in a region of space. Draw three equipotential surfaces in the region.
(b) Two point charges –q and +q are located at points (–a, 0, 0) and (a, 0, 0) respectively. Find the electrostatic potential at point (x, 0, 0) where x >> a.
(b) Two point charges are located at (-a, 0, 0) with charge -q and (a, 0, 0) with charge +q. The potential at point (x, 0, 0), where x >> a, is the algebraic sum of potentials due to each charge.
Potential V = k(-q) / |x + a| + k(q) / |x - a|, where k = 1 / (4 * pi * epsilon_0).
For x >> a, |x + a| approx x + a and |x - a| approx x - a.
Using binomial approximation: 1/(x + a) approx 1/x - a/x2 and 1/(x - a) approx 1/x + a/x2.
So, V = k[-q (1/x - a/x2) + q(1/x + a/x2)] = k[-q/x + q/x + a*q/x2 + a*q/x2] = k[2*a*q/x2].
Thus, potential at (x, 0, 0) for x >> a is V approx 2 * k * a * q / x2 which shows the potential decreases as 1/x2 far from the dipole.
Name the electromagnetic waves which are used for
(i) detection of fractures in bones
(ii) physiotherapy
(iii) radar systems
Also write their wavelength range.
(ii) Infrared rays are used in physiotherapy as they provide heat that helps in relaxing muscles and improving blood circulation.
(iii) Microwaves are used in radar systems due to their ability to travel long distances and reflect from objects, helping in detecting their position and speed.
(i) Define mutual inductance of a pair of coils. Write its SI unit.
(ii) A long solenoid of radius R and length L has n turns per unit length. A circular loop of radius r(<R) is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop.
Mutual inductance of a pair of coils is defined as the property by which the magnetic flux through one coil changes when the current in the other coil changes. It is the ratio of the magnetic flux linked with one coil to the current in the other coil producing it. Its SI unit is Henry (H).
(ii) Derivation of Mutual Inductance between Solenoid and Loop:
Let the solenoid have length L, radius R, and n turns per unit length. The magnetic field inside the long solenoid is B = mu_0 * n * I where I is the current.
Since the circular loop of radius r (where r < R) is placed at the center of the solenoid and both axes coincide, the magnetic flux through the loop due to current I in solenoid is:
Flux, Phi = B * Area_of_loop = mu_0 * n * I * pi * r2
Mutual inductance M is defined by Phi = M * I => M = Phi / I = mu_0 * n * pi * r2
Hence, the mutual inductance of the solenoid and the circular loop placed inside it is M = mu_0 * n * pi * r2.
Two long straight parallel conductors A and B carrying steady currents Iₐ and I_b in the same direction are separated by a distance d. Deduce the expression for the force acting on length L of conductor B due to conductor A and show in figure. Write the expression for the force acting on length L of conductor A due to conductor B and show that it follows Newton’s third law.
(b) Force on conductor A due to conductor B: Similarly, current I_b in conductor B produces magnetic field B_b at conductor A given by B_b = (mu_0 / 2pi) * (I_b / d). The force on length L of conductor A carrying current Iₐ is F_ab = Iₐ * L * B_b = (mu_0 / 2pi) * (I_b * Iₐ / d) * L. The direction of force on A is towards B.
Newton’s third law verification: The magnitudes of forces on A and B are equal, F_ab = F_ba, and they act in opposite directions. Hence, action and reaction forces are equal and opposite, confirming Newton’s third law.
This force causes the conductors to attract each other when currents are in the same direction.
(a) State Bohr’s second postulate and mention its significance.
(b) Prove that, in Bohr model of hydrogen atom as principal quantum number n becomes large, the energy levels get closer and closer.
(b) According to Bohr's model, the energy of electron in nth orbit is E_n = -13.6 eV / n2. The difference between adjacent energy levels is ΔE = E_n+1 - E_n = -13.6 (1/(n+1)2 - 1/n2). For large n, 1/(n+1)2 ≈ 1/n2 - 2/n3. Therefore ΔE ≈ 13.6 * 2/n3. As n increases, ΔE approaches zero, meaning energy levels get closer and closer.
(a) Explain the statement - “Current is a scalar although we represent current with an arrow.”
(b) Use Kirchhoff's rules to find the current through 3 Ω resistor in the circuit shown in the figure:
(b) To find the current through the 3 Ω resistor using Kirchhoff's rules, first assign current directions and label currents I1, I2, I3. Apply Kirchhoff's current law (junction rule) at the junction where currents meet to establish relations between currents, for example I3 = I1 + I2. Then apply Kirchhoff's voltage law (loop rule) to loops in the circuit to write equations based on the sum of voltage drops and emfs being zero. Solve these simultaneous equations to find the value of current through the 3 Ω resistor. This method ensures conservation of charge and energy in the circuit.
Derive an expression for the magnetic field 𝐁⃗, due to a circular coil of N turns, each of radius r carrying current I, at a distance ‘x’ from the centre along its axis.
Using the Biot-Savart law, the magnetic field dB due to a small element of current is proportional to Idl sinθ/r2. For a circular coil, the magnetic field at point P on its axis is given by:
B = (μ0 N I r2)/(2 (r2 + x2)3/2)
where μ0 is the permeability of free space (4π * 10-7 T m/A).
Explanation: Each turn contributes a magnetic field at point P; all fields add up because the coil is closely wound.
At x = 0 (the centre of the coil), the expression reduces to B = (μ0 N I)/(2r), which is the known formula for magnetic field at the centre of the coil.
Section E
(a) What are coherent sources? Why are they necessary for observing stable interference pattern? Draw a graph showing the variation of intensity of light with the position on the screen in Young’s double-slit experiment.
(b) Find the intensity of light at a point on the screen when two interfering waves of the same intensity (I₀) have a path difference of
(i) λ/4 and (ii) λ/3.
In Young’s double-slit experiment, the intensity of light on the screen varies with the position, producing bright and dark fringes due to constructive and destructive interference. The graph of intensity versus position shows maxima (bright fringes) and minima (dark fringes) alternately. The intensity I at any point on the screen is given by I = 4I₀ cos2(φ/2), where φ is the phase difference related to the path difference.
Graph: A sinusoidal curve starting from maximum intensity, representing intensity variation with position along the screen showing bright and dark fringes.
(b) Given two waves of the same intensity I₀ and path difference Δ = (i) λ/4, (ii) λ/3.
The resultant intensity I = 4I₀ cos2(φ/2), where φ = (2π/λ) * path difference.
(i) For Δ = λ/4, φ = 2π * (1/4) = π/2.
Thus I = 4I₀ cos2(π/4) = 4I₀ * (sqrt(2)/2)2 = 4I₀ * 1/2 = 2I₀.
(ii) For Δ = λ/3, φ = 2π * (1/3) = 2π/3.
Thus I = 4I₀ cos2(π/3) = 4I₀ * (1/2)2 = 4I₀ * 1/4 = I₀.
(a) Draw a labelled ray diagram of a refracting telescope when it forms image of a distant object at infinity. Derive expression for its magnifying power.
(b) (i) In a telescope the objective has much larger aperture than the eye piece. Why?
(ii) Write two advantages of reflecting telescope over refracting telescope.
Ray Diagram: Draw two convex lenses separated by the sum of their focal lengths (fo + fe). Parallel rays from a distant object strike the objective lens and converge to form an image at its focus. The eyepiece lens then makes the rays parallel again so the image is viewed at infinity.
Derivation of Magnifying Power (M): The angular magnification M is the ratio of the angle subtended by the image when viewed through the telescope to the angle subtended by the object when viewed with the naked eye.
Let angle subtended without telescope be θ and with telescope be θ'. Then M = θ' / θ.
For normal adjustment, the distance between the lenses is fo + fe.
Angular magnification M = fo / fe.
This means the magnifying power of the telescope is the ratio of the focal length of the objective lens to that of the eyepiece lens.
(b)(i) The objective lens has a much larger aperture than the eyepiece because a larger aperture collects more light from the distant object, making the image brighter and clearer. The eyepiece only magnifies the image and does not need to collect much light.
(b)(ii) Two advantages of reflecting telescope over refracting telescope are:
1. Reflecting telescopes do not suffer from chromatic aberration because mirrors reflect all colors equally, unlike lenses which disperse light.
2. They are easier and cheaper to make in large sizes because large mirrors are easier to support than large lenses, making them suitable for very powerful telescopes.
(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.
(b) Two air-filled capacitors of capacitances C₁ and C₂ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.
(b) Two capacitors of capacitances C₁ and C₂ are connected in parallel to a dc battery and fully charged. When a dielectric slab of constant K is inserted between the plates of each capacitor at constant voltage (due to battery), the capacitance of each capacitor increases by a factor K, so capacitance becomes K * C₁ and K * C₂.
(i) Since the capacitors remain connected to the battery, the voltage across each capacitor remains the same. Charge on each capacitor Q = C * V. So, after inserting the dielectric, charge on capacitor 1 becomes Q₁' = K * C₁ * V and on capacitor 2 becomes Q₂' = K * C₂ * V. Hence, charge on each capacitor increases by factor K.
(ii) Energy stored in a capacitor is given by U = 1/2 * C * V2. Since capacitance increases to K * C, and voltage remains constant, energy stored becomes U' = 1/2 * K * C * V2. Therefore, energy stored in each capacitor increases by factor K after dielectric insertion.
(a) An electric field 𝐄⃗ is established across the ends of a cylindrical conductor of length L and area of cross-section A. Discuss how electrons attain an average velocity, independent of time. Hence, obtain a relation between current in the conductor and this ‘average velocity’ of electrons.
(b) (i) This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?
(ii) Two copper wires having their radii in the ratio of 3 : 2 are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.
The current (I) in the conductor is given by I = nAe vd, where n is the number density of free electrons, e is the charge of an electron, A is the cross-sectional area, and vd is the average drift velocity.
(b)(i) Although the drift velocity is very slow (few mm/s), the electric field propagates through the conductor nearly at the speed of light. This rapid propagation causes electrons throughout the conductor to start drifting almost simultaneously, so current is established almost instantly when the circuit is closed.
(b)(ii) In series connection, current I is same through both wires. Let r1 and r2 be radii of wires with ratio 3:2.
Cross-sectional areas, A1 = π r12, A2 = π r22 so ratio A1 : A2 = 9 : 4.
Using I = n e A vd and same I, n, e, we get vd1 A1 = vd2 A2 => vd1/vd2 = A2/A1 = 4/9.
Thus, the ratio of drift velocities is 4 : 9.
(a) A series combination of L, C and R is connected to an a.c. source. Using a phasor diagram, derive an expression for the impedance of the circuit and phase difference between V and I.
(b) Under what conditions the
(i) impedance of the circuit is minimum ?
(ii) Wattless current flows in the circuit ?
The voltage across R is VR = I R, which is in phase with current I.
The voltage across inductor L is VL = I XL where XL = 2 pi f L, and it leads the current by 90deg.
The voltage across capacitor C is VC = I XC where XC = 1 / (2 pi f C), and it lags the current by 90deg.
In the phasor diagram, VR is along the reference axis (along I), VL is +90deg to I and VC is -90deg to I.
The net reactance voltage is VX = VL - VC = I (XL - XC).
Thus, the total voltage V = vector sum of VR and VX. The magnitude of voltage is V = I sqrt(R2 + (XL - XC)2).
Hence, the impedance Z = V / I = sqrt(R2 + (XL - XC)2).
Phase difference between voltage and current, phi = tan-1 ((XL - XC) / R). Voltage leads current if XL > XC and lags if XC > XL.
(b)
(i) The impedance of the circuit is minimum when the reactance terms cancel, i.e. XL = XC. This condition is called resonance.
(ii) Wattless current flows when the circuit current has zero phase difference with voltage across reactive elements, meaning current is purely reactive with no real power. This condition occurs when XL = XC, so circuit behaves like a pure resistance consuming no net power from the reactive parts.
(i) With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.
(ii) Deduce an expression for the induced emf in the coil of the generator.
(iii) If T is the time period of the rotation of the coil, at what values of T in a cycle, the emf generator is maximum ?
Construction: It consists of a rectangular coil ABCD mounted on an axle so that it can rotate freely between the poles of a magnet producing a magnetic field B. The coil is connected to slip rings R and R' which rotate with the coil. Carbon brushes B and B' press on the slip rings to conduct current to the external circuit.
Working: When the coil rotates, the magnetic flux linked with the coil changes continuously. According to Faraday's law, this change induces an emf in the coil. The direction of emf changes every half rotation, resulting in alternating current in the external circuit.
(ii) Expression for induced emf:
If N is the number of turns of the coil, A is the area of each turn, B is the magnetic field strength and θ is the angle between the coil's normal and magnetic field at time t, then magnetic flux ϕ = N * B * A * cosθ.
If the coil rotates with angular velocity ω, then θ = ωt.
Induced emf E = - dϕ/dt = - d/dt (N * B * A * cosωt) = N * B * A * ω * sinωt.
Therefore, E = Emax * sinωt where Emax = N * B * A * ω.
(iii) Maximum emf in a cycle:
Emf is maximum when sinωt = ±1, i.e., ωt = 90deg, 270deg, 450deg,...
Since ω = 2π/T, maximum emf occurs at t = T/4, 3T/4, 5T/4, etc., i.e., at odd multiples of T/4 in each cycle.
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