FYUG Even Semester Exam, 2025
Physics (2nd Semester)
Electricity and Magnetism (PHYDSC-151)

Paper Code: PHYDSC-151 | Full Marks: 50 | Time: As per Regulations


UNIT-I

Question 1 (a) [2]

What is an electric field line? Why two electric field lines cannot intersect each other?

An electric field line is an imaginary curve drawn in such a way that the tangent to it at any point gives the direction of the electric field intensity at that point.

Two electric field lines can never intersect because if they did, at the point of intersection, two tangents could be drawn. This would imply two different directions for the electric field at the same point, which is physically impossible.

Question 1 (b) [2]

Derive Poisson's equation from differential form of Gauss' law.

The differential form of Gauss' Law is given by:

grad . E = rho / epsilon_0

We know that the electric field E is the negative gradient of the electric potential V:

E = -grad V

Substituting E in Gauss' Law:

grad . (-grad V) = rho / epsilon_0

This results in Poisson's Equation:

grad²V = -rho / epsilon_0

Question 1 (c) [2]

Show that electrostatic field is conservative in nature.

A field is conservative if the work done in moving a charge between two points is independent of the path taken. Mathematically, the curl of the electric field must be zero:

curl E = 0

Since the line integral of E around any closed loop is zero (closed integral E.dl = 0), the energy is conserved, proving the field is conservative.

Question 2 (a) [10]

(i) State and prove Gauss' law. Apply Gauss' law to find the electric field due to a line charge.

Statement: Gauss' Law states that the total electric flux through any closed surface is equal to 1/epsilon_0 times the net charge enclosed by that surface.

Flux = closed integral E.dS = Q_enclosed / epsilon_0

Proof: Consider a point charge q at the center of a sphere of radius r. The field E is q / (4 * pi * epsilon_0 * r²). The area dS is 4 * pi * r². Multiplying these gives q / epsilon_0.

Electric Field due to Line Charge:

  • Consider an infinite line charge with linear charge density lambda.
  • Draw a cylindrical Gaussian surface of radius r and length L.
  • Flux passes only through the curved surface: E * (2 * pi * r * L).
  • By Gauss Law: E * (2 * pi * r * L) = (lambda * L) / epsilon_0.
E = lambda / (2 * pi * epsilon_0 * r)

(ii) Show that the function phi = x² - y² + z satisfies Laplace's equation.

Laplace's equation is grad²phi = 0.

  1. d(phi)/dx = 2x; d²(phi)/dx² = 2
  2. d(phi)/dy = -2y; d²(phi)/dy² = -2
  3. d(phi)/dz = 1; d²(phi)/dz² = 0

Summing them up: 2 + (-2) + 0 = 0. Thus, it satisfies the equation.

Question 2 (b) [OR Option] [10]

(i) What is electric dipole? Define dipole moment. Find the expression for potential and field intensity.

An electric dipole consists of two equal and opposite charges separated by a small distance. Dipole Moment (p) is the product of one charge and the distance between them: p = q * 2a.

Potential (V): At a point (r, theta), V = (p * cos(theta)) / (4 * pi * epsilon_0 * r²).

Field Intensity (E): E = [p * sqrt(1 + 3cos²(theta))] / (4 * pi * epsilon_0 * r³).

(ii) Find the expression for torque acting on an electric dipole.

When a dipole is placed in a uniform electric field E at an angle theta, the forces on the charges create a couple.

Torque (tau) = Force * perpendicular distance = (qE) * (2a sin(theta))
tau = p * E * sin(theta) OR tau = p x E

UNIT-II

Question 3 (a) [2]

What are meant by free charge and bound charge?

  • Free Charge: Charges that can move freely within a conductor (like conduction electrons).
  • Bound Charge: Charges that are restricted to specific atoms or molecules in a dielectric and can only shift slightly under an electric field.

Question 3 (b) [2]

Write down the relation connecting E, P and D. What is polarisation vector?

The relation is:

D = (epsilon_0 * E) + P

The polarisation vector (P) is defined as the electric dipole moment per unit volume of the dielectric material.

Question 3 (c) [2]

State Gauss' law in dielectrics.

The surface integral of the electric displacement field D over a closed surface is equal to the free charge enclosed by that surface: closed integral D.dS = Q_free.

Question 4 (a) [10]

(i) Define electrical susceptibility and dielectric constant. Obtain the relation between them.

Susceptibility (chi): Measures how easily a dielectric polarizes in an electric field: P = epsilon_0 * chi * E.
Dielectric Constant (K): The ratio of permittivity of the medium to the permittivity of vacuum.

Relation: Since D = epsilon_0 * E + P and D = epsilon_0 * K * E:
epsilon_0 * K * E = epsilon_0 * E + epsilon_0 * chi * E
Divide by (epsilon_0 * E):

K = 1 + chi

(ii) Find the expression for capacitance of a parallel plate capacitor with dielectric slab.

If a slab of thickness t and dielectric constant K is inserted between plates separated by d:

C = (epsilon_0 * A) / [d - t + (t/K)]

Question 4 (b) [OR Option] [10]

(i) Derive an expression for mechanical force per unit area on a charged conductor and find energy density.

The electrostatic pressure or force per unit area on the surface of a charged conductor is given by:

f = sigma² / (2 * epsilon_0)

Energy Density (u): The energy stored per unit volume in the electric field is:

u = (1/2) * epsilon_0 * E²

(ii) What is electrical image? Obtain electric field at a point on the plane surface.

The method of electrical images allows solving for the electric field of a charge near a conductor by replacing the conductor with a virtual "image charge". For a charge +q at distance d from a grounded plane, an image charge -q is placed at -d.

UNIT-III

Question 5 (a) [2]

Define magnetic field. What is the SI unit of magnetic field?

A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. The SI unit is the Tesla (T).

Question 5 (b) [2]

State Biot-Savart law.

It states that the magnetic field dB due to a current element I dl is proportional to the current, the length of the element, and the sine of the angle between the element and the position vector, and inversely proportional to the square of the distance.

dB = (mu_0 / 4pi) * (I dl sin(theta) / r²)

Question 6 (a) [10]

(i) Find magnetic field due to a current carrying circular loop at a point on the axis.

For a loop of radius R carrying current I, at a distance x from the center on the axis:

B = (mu_0 * I * R²) / [2 * (R² + x²)^(3/2)]

(ii) Show that div B = 0. What is its physical significance?

Since magnetic field lines form continuous closed loops, the net flux through any closed surface is zero. This is expressed as grad . B = 0. Physical Significance: It indicates that magnetic monopoles do not exist.

UNIT-IV

Question 7 (a) [2]

Neutral vs Inversion Temperature.

  • Neutral Temperature: The temperature of the hot junction at which the thermo-emf is maximum.
  • Inversion Temperature: The temperature at which the thermo-emf becomes zero and then reverses direction.

Question 8 (b) [10]

(i) Thomson effect and coefficient.

The Thomson effect is the evolution or absorption of heat when a current flows through a single conductor that has a temperature gradient. The Thomson coefficient (sigma) is the heat exchanged per unit current per unit temperature gradient.

(ii) EMF calculation for E = at + bt².

  • Neutral Temp (Tn): dE/dt = 0 -> a + 2bt = 0 -> Tn = -a/2b.
  • Inversion Temp (Ti): E = 0 -> t(a + bt) = 0 -> Ti = -a/b.
  • Peltier Coefficient (pi): pi = T * (dE/dT).

UNIT-V

Question 10 (b) [10]

(i) State and prove Norton's theorem.

Statement: Any linear bilateral network with voltage sources and resistances can be replaced by an equivalent circuit consisting of a single current source (In) in parallel with a single resistor (Rn).

(ii) Numerical: Unity power factor in AC circuit.

Given: C = 5 microFarads, R = 100 ohm, f = 50 Hz. For unity power factor, XL = XC.

XC = 1 / (2 * pi * f * C) = 1 / (2 * 3.14 * 50 * 5 * 10^-6) = 636.9 ohms
L = XC / (2 * pi * f) = 636.9 / (2 * 3.14 * 50) = 2.02 Henrys