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IGNOU MPH-003 Solved Assignment 2024 | MSCPH | IGNOU

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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Details For MPH-003 Solved Assignment

IGNOU MPH-003 Assignment Question Paper 2024

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mph-003-b0f6c82a-438d-4a97-8cd6-7d4a0723612d

  1. Two identical infinite non-conducting sheets having equal positive surface charge densities σ σ sigma\sigmaσ are kept parallel to each other as shown in the Figure below. Determine the electric field at a point in (a) region A A AAA on the left of the sheet 1, (b) region B B BBB between the sheets and (iii) region C C CCC on the right of the sheets.
original image
  1. Obtain the general solution of the two-dimensional Laplace’s equation in spherical polar coordinates given by
1 r 2 ∂ ∂ r ( r 2 ∂ V ∂ r ) + 1 r 2 sin ⁡ θ ∂ ∂ θ ( sin ⁡ θ ∂ V ∂ θ ) = 0 1 r 2 ∂ ∂ r r 2 ∂ V ∂ r + 1 r 2 sin ⁡ θ ∂ ∂ θ sin ⁡ θ ∂ V ∂ θ = 0 (1)/(r^(2))(del)/(del r)(r^(2)(del V)/(del r))+(1)/(r^(2)sin theta)(del)/(del theta)(sin theta(del V)/(del theta))=0\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial V}{\partial r}\right)+\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial V}{\partial \theta}\right)=01r2∂∂r(r2∂V∂r)+1r2sin⁡θ∂∂θ(sin⁡θ∂V∂θ)=0
  1. A point charge Q Q QQQ is situated at a distance D D DDD from the centre of an earthed conducting sphere of radius R R RRR where D > R D > R D > RD>RD>R. Using the method of images, determine the value and position of image charge and calculate the potential and electric field at a point outside the sphere.
  2. Considering a simple physical model of a dielectric as an aggregate of a large number of simple harmonic oscillators, obtain an expression for the dielectric constant.
  3. Using the multipole expansion technique, obtain the expression for the magnetic vector potential due to a localized current distribution at a distant point.
  4. Calculate the magnetic field inside and outside of a very long solenoid consisting n n nnn turns per unit length on a cylinder of radius R R RRR and carrying a steady current I I III.
  5. Obtain an expression for torque on a current loop kept in a uniform magnetic field.
  6. A toroid has mean circumference 0.4 m 0.4 m 0.4m0.4 \mathrm{~m}0.4 m and has 600 turns, each carrying a current of 0.12 A. (a) Calculate H → H → vec(H)\vec{H}H→ and B → B → vec(B)\vec{B}B→ if the toroid has an air core. (b) Calculate B → B → vec(B)\vec{B}B→ and the magnetisation M → M → vec(M)\vec{M}M→ if the core is filled with iron of relative permeability 4000 .
\(c=a\:cos\:B+b\:cos\:A\)

MPH-003 Sample Solution 2024

mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

  1. Two identical infinite non-conducting sheets having equal positive surface charge densities σ σ sigma\sigmaσ are kept parallel to each other as shown in the Figure below. Determine the electric field at a point in (a) region A A AAA on the left of the sheet 1, (b) region B B BBB between the sheets and (iii) region C C CCC on the right of the sheets.
original image
Answer:
We need to consider the superposition principle, which states that the net electric field is the vector sum of electric fields produced by each charge distribution. For an infinite sheet with a uniform positive surface charge density σ σ sigma\sigmaσ, the electric field at any point in space is perpendicular to the surface of the sheet and has a magnitude of E = σ 2 ϵ 0 E = σ 2 ϵ 0 E=(sigma)/(2epsilon_(0))E = \frac{\sigma}{2\epsilon_0}E=σ2ϵ0, according to Gauss’s law.
Let’s consider the situation described, with two identical infinite non-conducting sheets parallel to each other, each with a surface charge density of σ σ sigma\sigmaσ.
(a) Region A (to the left of sheet 1):
Since sheet 1 has a positive surface charge density σ σ sigma\sigmaσ, it creates an electric field directed away from itself. By symmetry, the electric field at any point to the left of sheet 1 will be directed to the right, perpendicular to the sheet, with a magnitude of σ 2 ϵ 0 σ 2 ϵ 0 (sigma)/(2epsilon_(0))\frac{\sigma}{2\epsilon_0}σ2ϵ0.
The electric field contribution from sheet 2 to the left of sheet 1 is exactly canceled by the field due to sheet 1, because the field from sheet 2 at that point is directed to the left, while the field from sheet 1 is to the right. Thus, the net electric field in region A is solely due to sheet 1 and is given by:
E A = σ 2 ϵ 0 E A = σ 2 ϵ 0 E_(A)=(sigma)/(2epsilon_(0))E_A = \frac{\sigma}{2\epsilon_0}EA=σ2ϵ0
(b) Region B (between the sheets):
In this region, both sheets contribute to the electric field. The electric field due to sheet 1 is directed to the right, and the electric field due to sheet 2 is also directed to the right, away from sheet 2. Since the sheets are identical, each contributes an electric field with a magnitude of σ 2 ϵ 0 σ 2 ϵ 0 (sigma)/(2epsilon_(0))\frac{\sigma}{2\epsilon_0}σ2ϵ0. These fields add up due to superposition.
The net electric field in region B is the sum of the fields due to both sheets:
E B = E sheet 1 + E sheet 2 = σ 2 ϵ 0 + σ 2 ϵ 0 = σ ϵ 0 E B = E sheet 1 + E sheet 2 = σ 2 ϵ 0 + σ 2 ϵ 0 = σ ϵ 0 E_(B)=E_(“sheet 1”)+E_(“sheet 2”)=(sigma)/(2epsilon_(0))+(sigma)/(2epsilon_(0))=(sigma)/(epsilon_(0))E_B = E_{\text{sheet 1}} + E_{\text{sheet 2}} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}EB=Esheet 1+Esheet 2=σ2ϵ0+σ2ϵ0=σϵ0
(iii) Region C (to the right of sheet 2):
For region C, the analysis is similar to region A. The electric field due to sheet 1 is canceled by the field due to sheet 2, and the net electric field in region C is solely due to sheet 2.
The electric field in region C, similar to region A, is directed away from sheet 2, to the right, with a magnitude of:
E C = σ 2 ϵ 0 E C = σ 2 ϵ 0 E_(C)=(sigma)/(2epsilon_(0))E_C = \frac{\sigma}{2\epsilon_0}EC=σ2ϵ0
In summary:
  • The electric field in region A (left of sheet 1) is E A = σ 2 ϵ 0 E A = σ 2 ϵ 0 E_(A)=(sigma)/(2epsilon_(0))E_A = \frac{\sigma}{2\epsilon_0}EA=σ2ϵ0, directed to the right.
  • The electric field in region B (between the sheets) is E B = σ ϵ 0 E B = σ ϵ 0 E_(B)=(sigma)/(epsilon_(0))E_B = \frac{\sigma}{\epsilon_0}EB=σϵ0, also directed to the right.
  • The electric field in region C (right of sheet 2) is E C = σ 2 ϵ 0 E C = σ 2 ϵ 0 E_(C)=(sigma)/(2epsilon_(0))E_C = \frac{\sigma}{2\epsilon_0}EC=σ2ϵ0, directed to the right.
This analysis is consistent with the principles of superposition and the symmetry of the electric field due to an infinite charged sheet.
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