IGNOU BPHCT-133 Solved Question Paper PDF Download

IGNOU BPHCT-133 Previous Year Solved Question Paper in Hindi

IGNOU BPHCT-133 Previous Year Solved Question Paper in English

Q1. Answer any five parts : 5×3=15 (a) Determine the unit vector normal to the curve : x³y⁴ + 6y² = x at the point (1, 0). (b) Calculate the work done by a force field F = xyî + y²ĵ in moving an object along the curve y=x² in the xy-plane from (0,0) to (1,1). (c) A point charge is enclosed by a spherical Gaussian surface. Would the electric flux through the surface change (i) if the Gaussian surface is chosen to be a closed cylinder or a cube ? (ii) if the charge is moved outside the Gaussian surface ? (iii) if another charge is placed inside the Gaussian surface ? (d) The electric field of an electromagnetic wave in vacuum is given by: Eₓ = 0, Eᵧ = 30cos(π/3 x – 2π×10⁶t), E₂=0 where Eᵧ is in Vm⁻¹, t in s and x in m. Determine the frequency, wavelength, the direction of propagation of the wave and the direction of the associated magnetic field. (e) Two point charges +q and -2q are placed along a straight line at a distance of 9 m from each other. Determine the distance of a point from the charge +q between the two charges where the electric potential is zero. (f) A thin dielectric rod of cross-section A extends along the x-axis from x=0 to x=L. The polarization of the rod is along its length and is given by P = (ax² + b)î. Obtain the bound volume charge density and the surface charge density at each end of the rod. (g) A long cylindrical wire of radius R carries a steady current i which is uniformly distributed over its cross-sectional area. Determine the magnetic field at a distance r(<R) from the axis of the wire. (h) Obtain the maximum value of the displacement current in a parallel plate capacitor made up of plates of area A. The electric field between the plates is given by E = E₀ sin(ωt).

Ans. (a) The curve is defined by the equation `f(x, y) = x³y⁴ + 6y² – x = 0`. The vector normal to the curve at any point (x, y) is given by the gradient of the scalar function `f(x, y)`, which is `∇f`. `∇f = (∂f/∂x)î + (∂f/∂y)ĵ` Here, `∂f/∂x = 3x²y⁴ – 1` and `∂f/∂y = 4x³y³ + 12y`. At the point (1, 0): `∂f/∂x` at (1,0) = `3(1)²(0)⁴ – 1 = -1` `∂f/∂y` at (1,0) = `4(1)³(0)³ + 12(0) = 0` So, the normal vector `N` at (1, 0) is `∇f = -1î + 0ĵ = -î`. The unit vector normal `n̂` is calculated as: `n̂ = N / |N| = (-î) / √((-1)² + 0²) = -î / 1 = -î` Therefore, the unit vector normal at the point (1, 0) is -î .

(b) The work done `W` by the force field `F = xyî + y²ĵ` is given by the line integral `W = ∫ F · dr`. The curve is `y = x²`, so `dy = 2x dx`. The displacement vector is `dr = dx î + dy ĵ = dx î + 2x dx ĵ`. `F · dr = (xyî + y²ĵ) · (dx î + 2x dx ĵ) = (xy + 2xy²) dx` Now, substituting the curve’s equation `y = x²` into the expression: `F · dr = (x(x²) + 2x(x²)²) dx = (x³ + 2x⁵) dx` To find the work done, we integrate from x=0 to x=1: `W = ∫₀¹ (x³ + 2x⁵) dx = [x⁴/4 + 2x⁶/6]₀¹` `W = [x⁴/4 + x⁶/3]₀¹ = (1⁴/4 + 1⁶/3) – (0) = 1/4 + 1/3 = (3+4)/12 = 7/12` Thus, the work done is 7/12 Joules.

(c) According to Gauss’s Law, the total electric flux through a closed surface is `Φ = q_enclosed / ε₀`, where `q_enclosed` is the total charge enclosed by the surface.

• (i) No, the electric flux would not change. Gauss’s Law states that the flux depends only on the amount of charge enclosed , not on the size or shape of the Gaussian surface. Therefore, the flux remains the same whether the surface is a sphere, cylinder, or cube.
• (ii) Yes, the electric flux would change. If the charge is moved outside the Gaussian surface, the enclosed charge `q_enclosed` becomes zero. Therefore, the flux through the surface will become zero .
• (iii) Yes, the electric flux would change. If another charge is placed inside the Gaussian surface, the total enclosed charge `q_enclosed` will change. Consequently, the total electric flux through the surface will also change.

(d) Comparing the given electric field `Eᵧ = 30cos(π/3 x – 2π×10⁶t)` with the standard wave equation `E = E₀cos(kx – ωt)`: We have `k = π/3 rad/m` and `ω = 2π × 10⁶ rad/s`.

• Frequency (f): `ω = 2πf` ⇒ `f = ω / 2π = (2π × 10⁶) / 2π = 10⁶ Hz` or 1 MHz.
• Wavelength (λ): `k = 2π/λ` ⇒ `λ = 2π / k = 2π / (π/3) = 6 m`.
• Direction of propagation: The argument is of the form `(kx – ωt)`, which indicates the wave is propagating in the +x direction .
• Direction of magnetic field: In an EM wave, `E`, `B`, and the direction of propagation `k` are mutually perpendicular. `E` is in the +y direction (`ĵ`) and propagation is in the +x direction (`î`). From the relation `k̂ ∝ Ê × B̂`, we have `î ∝ ĵ × B̂`. For this to be true, `B̂` must be in the +z direction (`k̂`).

(e) Let the point where the electric potential is zero be at a distance `x` from the charge `+q`. This point is between the two charges. The distance from the charge `-2q` will be `(9 – x)` meters. The total potential `V` at a point is the scalar sum of potentials due to each charge: `V = V₁ + V₂`. `V = k(+q)/x + k(-2q)/(9-x)` For zero potential, `V = 0`. `kq/x – 2kq/(9-x) = 0` `1/x = 2/(9-x)` `9 – x = 2x` `3x = 9` `x = 3` meters. Thus, the point is at a distance of 3 meters from the charge `+q`.

(f) The given polarization is `P = (ax² + b)î`. The bound volume charge density is given by `ρ_b = -∇ · P`. `ρ_b = – d/dx (ax² + b) = -2ax` So, the bound volume charge density `ρ_b = -2ax` . The bound surface charge density is given by `σ_b = P · n̂`, where `n̂` is the outward normal.

• At the end `x=0`, the outward normal is `n̂ = -î`. `σ_b(x=0) = P(x=0) · (-î) = (a(0)² + b)î · (-î) = b(-1) = -b`.
• At the end `x=L`, the outward normal is `n̂ = +î`. `σ_b(x=L) = P(x=L) · (+î) = (a(L)² + b)î · (+î) = aL² + b`.

Thus, the

surface charge densities are `σ_b(0) = -b`

and

`σ_b(L) = aL² + b`

at the respective ends.

(g) To find the magnetic field inside the wire (`r < R`), we use Ampere’s Law: `∮ B · dl = μ₀ I_enclosed`. We choose a circular Amperian loop of radius `r` concentric with the wire’s axis. `∮ B · dl = B * (2πr)` Since the current is uniformly distributed, the current density is `J = i / (πR²)`. The current enclosed within the radius `r`, `I_enclosed`, is: `I_enclosed = J × (Area of loop) = (i / πR²) × (πr²) = i(r²/R²)` Substituting into Ampere’s law: `B(2πr) = μ₀ * i(r²/R²)` `B = (μ₀ i r) / (2πR²)` Thus, the magnetic field inside the wire is `B = (μ₀ i r) / (2πR²)` , which increases linearly with the distance `r` from the axis.

(h) The displacement current is given by `I_d = ε₀ (dΦ_E / dt)`, where `Φ_E` is the electric flux. For a parallel plate capacitor, the electric field `E` is assumed to be uniform between the plates. `Φ_E = E · A = E A` (since E and A are parallel). Given `E = E₀ sin(ωt)`. So, `Φ_E = (E₀ sin(ωt)) A = E₀A sin(ωt)`. Now, calculate `I_d`: `I_d = ε₀ d/dt (E₀A sin(ωt)) = ε₀ E₀ A d/dt(sin(ωt))` `I_d = ε₀ E₀ A ω cos(ωt)` The maximum value of the displacement current occurs when `cos(ωt)` is maximum, which is 1. Therefore, the maximum displacement current is: `I_d(max) = ε₀ E₀ A ω` .

Q2. Answer any five parts : 5×5=25 (a) An AC generator consists of 20 turns square wire coil of side 50 cm. The coil is turned at 50 revolutions per second to produce the standard 50 Hz alternating current produced in the country. What must the magnitude of the magnetic field be for the peak output voltage of the generator to be 300 V ? (b) Five very long straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are i₁=20A, i₂=-6A, i₃=12A, i₄=-7A and i₅=18A (negative currents are opposite in direction to the positive). Calculate the magnitude of B at a distance of 10 cm from the cable. (c) Calculate the magnitude of the magnetic force exerted by Earth’s magnetic field B = 1.0×10⁻⁷ T on an electron moving with speed 1.0×10⁵ ms⁻¹ near the Earth’s surface. (d) Write Maxwell’s equations in differential form. Derive the wave equation for B field. (e) Consider a system of four charges : 4q, q, -4q and -q. Draw a Gaussian surface enclosing at least two charges of the system so that the net electric flux through it is (i) zero, (ii) q/ε₀. (f) A charge of 2.0μC is kept at the origin and points A and B are located along the x-axis at a distance of 10 cm and 50 cm from the origin. Calculate the work done in bringing a charge of 1.0μC from point B to A. (g) Using the Divergence Theorem, calculate the flux of a vector field : F = z²î + 4yĵ – x²k̂ over a sphere of radius 1 unit. (h) A central force field is a force field of the form F = f(r)r̂. Determine ∇ × F.

Ans. (a) Given values are: Number of turns, `N = 20` Side of the coil, `l = 50 cm = 0.5 m` Area of the coil, `A = l² = (0.5)² = 0.25 m²` Frequency, `f = 50 Hz` Angular frequency, `ω = 2πf = 2π(50) = 100π rad/s` Peak voltage, `ε_max = 300 V` The formula for the peak voltage (emf) in an AC generator is: `ε_max = N A B ω` We need to find the magnitude of the magnetic field, B: `B = ε_max / (N A ω)` `B = 300 / (20 × 0.25 × 100π) = 300 / (5 × 100π) = 300 / (500π)` `B = 3 / (5π) ≈ 3 / (5 × 3.14159) ≈ 0.191 T` Thus, the magnitude of the required magnetic field is approximately 0.191 Tesla .

(b) By Ampere’s law, for a point outside the closely bound cable, the magnetic field is the same as that of a single wire carrying the net current. First, calculate the net current `I_net`: `I_net = i₁ + i₂ + i₃ + i₄ + i₅ = 20 – 6 + 12 – 7 + 18 = 37 A` The magnitude of the magnetic field `B` at a distance `r` from a long, straight wire is given by: `B = μ₀ I_net / (2πr)` Here, `r = 10 cm = 0.1 m` and `μ₀ = 4π × 10⁻⁷ T·m/A`. `B = (4π × 10⁻⁷ T·m/A × 37 A) / (2π × 0.1 m)` `B = (2 × 10⁻⁷ × 37) / 0.1 = 74 × 10⁻⁷ T = 7.4 × 10⁻⁶ T` The magnitude of the magnetic field at 10 cm from the cable is 7.4 × 10⁻⁶ T .

(c) The magnetic force (Lorentz force) on a moving charge in a magnetic field is `F = q(v × B)`. Its magnitude is `F = |q|vBsinθ`, where `θ` is the angle between the velocity `v` and the magnetic field `B`. The maximum force occurs when `sinθ = 1` (i.e., `θ = 90°`). Given values: Charge of electron, `|q| = e = 1.6 × 10⁻¹⁹ C` Speed of electron, `v = 1.0 × 10⁵ m/s` Magnetic field, `B = 1.0 × 10⁻⁷ T` The magnitude of the maximum force will be: `F_max = |q|vB = (1.6 × 10⁻¹⁹ C) × (1.0 × 10⁵ m/s) × (1.0 × 10⁻⁷ T)` `F_max = 1.6 × 10⁻¹⁹⁺⁵⁻⁷ N = 1.6 × 10⁻²¹ N` Thus, the magnitude of the magnetic force on the electron is 1.6 × 10⁻²¹ N .

(d) Maxwell’s Equations (Differential Form): In vacuum (with no free charges or currents), the four Maxwell’s equations are:

1. Gauss’s Law for Electricity: `∇ · E = 0`
2. Gauss’s Law for Magnetism: `∇ · B = 0`
3. Faraday’s Law of Induction: `∇ × E = -∂B/∂t`
4. Ampere-Maxwell Law: `∇ × B = μ₀ε₀ ∂E/∂t`

Derivation of Wave Equation for B-field:

1. Take the curl of the Ampere-Maxwell Law: `∇ × (∇ × B) = ∇ × (μ₀ε₀ ∂E/∂t) = μ₀ε₀ ∂/∂t (∇ × E)`.

2. Use the vector identity `∇ × (∇ × B) = ∇(∇ · B) – ∇²B`.

3. From Gauss’s Law for Magnetism (`∇ · B = 0`), this simplifies to `∇ × (∇ × B) = -∇²B`.

4. Substitute Faraday’s Law (`∇ × E = -∂B/∂t`) into the right side of the equation from step 1: `μ₀ε₀ ∂/∂t (-∂B/∂t) = -μ₀ε₀ ∂²B/∂t²`.

5. Equate the results from steps 3 and 4: `-∇²B = -μ₀ε₀ ∂²B/∂t²`.

`∇²B = μ₀ε₀ ∂²B/∂t²`

This is the

wave equation

for the B-field, which can also be written as `∇²B = (1/c²) ∂²B/∂t²` where `c = 1/√(μ₀ε₀)`.

(e) From Gauss’s Law, the electric flux is `Φ = q_enclosed / ε₀`. (i) Zero Flux: For this, `q_enclosed` must be zero. We can draw a Gaussian surface that encloses charges that sum to zero. For example, a surface enclosing `q` and `-q`, or a surface enclosing `4q` and `-4q`. Diagram: A surface is drawn enclosing the charges `q` and `-q`. Total enclosed charge = `q + (-q) = 0`. Thus, the flux is zero. (ii) Flux q/ε₀: For this, `q_enclosed = q`. Since we must enclose at least two charges, we can draw a surface that encloses the charges `4q`, `-4q`, and `q`. Diagram: A surface is drawn enclosing the charges `4q`, `-4q`, and `q`. Total enclosed charge = `4q + (-4q) + q = q`. Thus, the flux is `q/ε₀`.

(f) The work done in moving a test charge `q_test` from point B to point A is `W = q_test (V_A – V_B)`. Given values: Source charge, `q_source = 2.0 μC = 2.0 × 10⁻⁶ C` Test charge, `q_test = 1.0 μC = 1.0 × 10⁻⁶ C` Distance to point A, `r_A = 10 cm = 0.1 m` Distance to point B, `r_B = 50 cm = 0.5 m` The potential `V = k * q_source / r`, where `k = 9 × 10⁹ N·m²/C²`. `V_A = (9 × 10⁹ × 2.0 × 10⁻⁶) / 0.1 = 18 × 10³ / 0.1 = 1.8 × 10⁵ V` `V_B = (9 × 10⁹ × 2.0 × 10⁻⁶) / 0.5 = 18 × 10³ / 0.5 = 0.36 × 10⁵ V` Calculate the work done `W`: `W = (1.0 × 10⁻⁶ C) × (1.8 × 10⁵ V – 0.36 × 10⁵ V)` `W = 1.0 × 10⁻⁶ × (1.44 × 10⁵) J = 0.144 J` Therefore, the work done is 0.144 J .

(g) The Divergence Theorem states that the flux `Φ` of a vector field `F` through a closed surface `S` is equal to the volume integral of the divergence of `F` over the volume `V` enclosed by the surface: `Φ = ∬_S F · dS = ∭_V (∇ · F) dV`. Given vector field: `F = z²î + 4yĵ – x²k̂`. The divergence of `F` is: `∇ · F = ∂/∂x(z²) + ∂/∂y(4y) + ∂/∂z(-x²) = 0 + 4 + 0 = 4` The flux is: `Φ = ∭_V (4) dV = 4 ∭_V dV` The integral `∭_V dV` is the volume of the sphere. Given the radius is 1 unit. Volume of sphere, `V = (4/3)πr³ = (4/3)π(1)³ = 4π/3` `Φ = 4 × (4π/3) = 16π/3` Thus, the flux over the sphere is 16π/3 .

(h) A central force field has the form `F = f(r)r̂`, where `r = |r|` and `r̂ = r/r`. We can write `r = xî + yĵ + zk̂` and `r = √(x² + y² + z²)`. `F = f(r) * (xî + yĵ + zk̂) / r`. Let `g(r) = f(r)/r`, so `F = g(r)x î + g(r)y ĵ + g(r)z k̂`. We need to calculate `∇ × F`. The x-component of the curl is: `(∇ × F)ₓ = ∂/∂y(Fz) – ∂/∂z(Fy) = ∂/∂y(g(r)z) – ∂/∂z(g(r)y)` Using the chain rule: `∂(g(r))/∂y = (dg/dr)(∂r/∂y) = g'(r)(y/r)`. `(∇ × F)ₓ = z g'(r)(y/r) – y g'(r)(z/r) = (zy/r)g'(r) – (yz/r)g'(r) = 0` By symmetry, the y and z components will also be zero. `(∇ × F)ᵧ = ∂/∂z(Fx) – ∂/∂x(Fz) = 0` `(∇ × F)₂ = ∂/∂x(Fy) – ∂/∂y(Fx) = 0` Therefore, for any central force field, `∇ × F = 0`. This means that central force fields are always irrotational and therefore conservative .

Q3. Answer any one part: 1×10=10 (a) Determine the electric potential V of a uniformly charged non-conducting sphere of radius R at a point P located at a distance ‘r’ from the centre of the sphere (r<R). (b) State Biot-Savart’s law. Determine the magnetic field B at a point P located at a distance R along the axis of a circular current loop carrying a current I.

Ans. (a) We consider a non-conducting sphere of radius `R` and total charge `Q` uniformly distributed throughout its volume. The charge density is `ρ = Q / ((4/3)πR³)`. To find the potential `V`, we first need the electric field `E`. Using Gauss’s law, we find the electric field inside and outside the sphere:

• Outside the sphere (`r > R`): `E_out = (1 / 4πε₀) * (Q / r²)`
• Inside the sphere (`r < R`): `E_in = (1 / 4πε₀) * (Qr / R³)`

The potential `V` is defined as `V(r) = -∫ E · dr`. We take the reference point `V(∞)=0`. To find the potential at a point inside the sphere (`r < R`), we integrate the work done in bringing a charge from infinity to that point, which can be split into two parts:

`V(r) = – ∫[from ∞ to R] E_out dr – ∫[from R to r] E_in dr`

The first integral gives the potential at the surface, `V(R)`:

`V(R) = – ∫[from ∞ to R] (kQ/r²) dr = [-(-kQ/r)] from ∞ to R = kQ/R` (where `k = 1/4πε₀`)

Now, we integrate the second part:

`-∫[from R to r] E_in dr = -∫[from R to r] (kQr/R³) dr`

`= -(kQ/R³) [r²/2] from R to r`

`= -(kQ/R³) (r²/2 – R²/2) = -kQr²/(2R³) + kQR²/(2R³)`

`= -kQr²/(2R³) + kQ/(2R)`

So the total potential at a point `r` inside is:

`V(r) = V(R) – ∫[from R to r] E_in dr`

`V(r) = kQ/R + [-kQr²/(2R³) + kQ/(2R)]`

`V(r) = (3kQ/2R) – (kQr² / 2R³)`

This can be rewritten as:

`V(r) = (kQ / 2R³) * (3R² – r²)`

Or

`V(r) = (Q / 8πε₀R³) * (3R² – r²)`

This is the expression for the

electric potential

at a point at distance `r` from the center of a uniformly charged non-conducting sphere of radius `R`.

(b) Biot-Savart’s Law: Biot-Savart’s law is an equation that describes the magnetic field generated by a constant electric current. It states that the magnetic field `dB` produced by an infinitesimal current element `Idl` at a point P, located at a position vector `r` from the element, is given by: `dB = (μ₀ / 4π) * (I dl × r̂) / r²` where:

• `μ₀` is the permeability of free space.
• `I` is the current in the wire.
• `dl` is the vector of the current element, with its direction being the direction of the current.
• `r` is the distance from the current element to the point P.
• `r̂` is the unit vector pointing from the current element to the point P.

The direction of `dB` is perpendicular to both `dl` and `r̂`, given by the right-hand rule.

Magnetic Field on the Axis of a Circular Current Loop: Consider a circular loop of radius `a` carrying a current `I`. We want to find the magnetic field `B` at a point P on the axis of the loop, at a distance `R` from the center. 1. Consider a small current element `dl` on the loop. The distance from this element to point P is `r = √(a² + R²)`. 2. According to the Biot-Savart law, the magnitude of the magnetic field `dB` at P due to this element is: `dB = (μ₀ / 4π) (I dl / r²) = (μ₀ / 4π) (I dl / (a² + R²))`। 3. The direction of `dB` is perpendicular to both the `dl` vector and the vector `r`. We can resolve `dB` into two components: one along the axis (`dB_axial`) and one perpendicular to the axis (`dB_perp`). 4. By symmetry, for every `dl` on the loop, there is a diametrically opposite element whose perpendicular component (`dB_perp`) cancels the first one. Therefore, all perpendicular components cancel out. 5. Only the axial components (`dB_axial`) add up. If `θ` is the angle between the axis and the vector `r`, then `dB_axial = dB sin(θ)`. From the geometry, `sin(θ) = a / r = a / √(a² + R²)`. `dB_axial = [(μ₀ / 4π) (I dl / (a² + R²))] [a / √(a² + R²)]` `dB_axial = (μ₀ I a / 4π) * (dl / (a² + R²)^(3/2))` 6. To get the total magnetic field `B`, we integrate `dB_axial` over the entire loop: `B = ∫ dB_axial = ∫ (μ₀ I a / (4π(a² + R²)^(3/2))) dl` Since `μ₀, I, a, R` are constants, they come out of the integral: `B = (μ₀ I a / (4π(a² + R²)^(3/2))) ∫ dl` The integral `∫dl` is the circumference of the loop, which is `2πa`. `B = (μ₀ I a / (4π(a² + R²)^(3/2))) * (2πa)` `B = (μ₀ I a²) / (2(a² + R²)^(3/2))` This is the expression for the magnetic field at a point on the axis of a circular current loop. Its direction is along the axis, determined by the right-hand thumb rule.