# IGNOU MPH-003 Solved Assignment 2024 | MSCPH | IGNOU

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

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## IGNOU MPH-003 Assignment Question Paper 2024

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

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

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

## 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 $\sigma$$\sigma$sigma\sigma$\sigma$ are kept parallel to each other as shown in the Figure below. Determine the electric field at a point in (a) region $A$$A$AA$A$ on the left of the sheet 1, (b) region $B$$B$BB$B$ between the sheets and (iii) region $C$$C$CC$C$ on the right of the sheets.
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 $\sigma$$\sigma$sigma\sigma$\sigma$, the electric field at any point in space is perpendicular to the surface of the sheet and has a magnitude of $E=\frac{\sigma }{2{ϵ}_{0}}$$E=\frac{\sigma }{2{ϵ}_{0}}$E=(sigma)/(2epsilon_(0))E = \frac{\sigma}{2\epsilon_0}$E=\frac{\sigma }{2{ϵ}_{0}}$, 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 $\sigma$$\sigma$sigma\sigma$\sigma$.
(a) Region A (to the left of sheet 1):
Since sheet 1 has a positive surface charge density $\sigma$$\sigma$sigma\sigma$\sigma$, 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 $\frac{\sigma }{2{ϵ}_{0}}$$\frac{\sigma }{2{ϵ}_{0}}$(sigma)/(2epsilon_(0))\frac{\sigma}{2\epsilon_0}$\frac{\sigma }{2{ϵ}_{0}}$.
The electric field contribution from sheet 2 to the left of sheet 1 is exactly canceled by the field due to sheet 1, because the field from sheet 2 at that point is directed to the left, while the field from sheet 1 is to the right. Thus, the net electric field in region A is solely due to sheet 1 and is given by:
${E}_{A}=\frac{\sigma }{2{ϵ}_{0}}$${E}_{A}=\frac{\sigma }{2{ϵ}_{0}}$E_(A)=(sigma)/(2epsilon_(0))E_A = \frac{\sigma}{2\epsilon_0}${E}_{A}=\frac{\sigma }{2{ϵ}_{0}}$
(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 $\frac{\sigma }{2{ϵ}_{0}}$$\frac{\sigma }{2{ϵ}_{0}}$(sigma)/(2epsilon_(0))\frac{\sigma}{2\epsilon_0}$\frac{\sigma }{2{ϵ}_{0}}$. These fields add up due to superposition.
The net electric field in region B is the sum of the fields due to both sheets:
${E}_{B}={E}_{\text{sheet 1}}+{E}_{\text{sheet 2}}=\frac{\sigma }{2{ϵ}_{0}}+\frac{\sigma }{2{ϵ}_{0}}=\frac{\sigma }{{ϵ}_{0}}$${E}_{B}={E}_{\text{sheet 1}}+{E}_{\text{sheet 2}}=\frac{\sigma }{2{ϵ}_{0}}+\frac{\sigma }{2{ϵ}_{0}}=\frac{\sigma }{{ϵ}_{0}}$E_(B)=E_(“sheet 1”)+E_(“sheet 2”)=(sigma)/(2epsilon_(0))+(sigma)/(2epsilon_(0))=(sigma)/(epsilon_(0))E_B = E_{\text{sheet 1}} + E_{\text{sheet 2}} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}${E}_{B}={E}_{\text{sheet 1}}+{E}_{\text{sheet 2}}=\frac{\sigma }{2{ϵ}_{0}}+\frac{\sigma }{2{ϵ}_{0}}=\frac{\sigma }{{ϵ}_{0}}$
(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}=\frac{\sigma }{2{ϵ}_{0}}$${E}_{C}=\frac{\sigma }{2{ϵ}_{0}}$E_(C)=(sigma)/(2epsilon_(0))E_C = \frac{\sigma}{2\epsilon_0}${E}_{C}=\frac{\sigma }{2{ϵ}_{0}}$
In summary:
• The electric field in region A (left of sheet 1) is ${E}_{A}=\frac{\sigma }{2{ϵ}_{0}}$${E}_{A}=\frac{\sigma }{2{ϵ}_{0}}$E_(A)=(sigma)/(2epsilon_(0))E_A = \frac{\sigma}{2\epsilon_0}${E}_{A}=\frac{\sigma }{2{ϵ}_{0}}$, directed to the right.
• The electric field in region B (between the sheets) is ${E}_{B}=\frac{\sigma }{{ϵ}_{0}}$${E}_{B}=\frac{\sigma }{{ϵ}_{0}}$E_(B)=(sigma)/(epsilon_(0))E_B = \frac{\sigma}{\epsilon_0}${E}_{B}=\frac{\sigma }{{ϵ}_{0}}$, also directed to the right.
• The electric field in region C (right of sheet 2) is ${E}_{C}=\frac{\sigma }{2{ϵ}_{0}}$${E}_{C}=\frac{\sigma }{2{ϵ}_{0}}$E_(C)=(sigma)/(2epsilon_(0))E_C = \frac{\sigma}{2\epsilon_0}${E}_{C}=\frac{\sigma }{2{ϵ}_{0}}$, 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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