BE Computer Engineering (IOE, TU) Engineering Physics (IOE, SH 452) Question Paper 2078 Nepal

(a) Damped harmonic oscillator [7 marks]

A particle of mass $m$ experiences a restoring force $-kx$ and a damping force proportional to velocity $-b\dfrac{dx}{dt}$. Newton's second law gives:

$$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}$$

Dividing by $m$ and writing $2\beta = b/m$ (damping constant) and $\omega_0^2 = k/m$ (natural frequency):

$$\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = 0$$

Solution. Try $x = e^{\alpha t}$. The auxiliary equation is

$$\alpha^2 + 2\beta\alpha + \omega_0^2 = 0 \quad\Rightarrow\quad \alpha = -\beta \pm \sqrt{\beta^2 - \omega_0^2}$$

Underdamped case ($\beta < \omega_0$): the root is complex, $\alpha = -\beta \pm i\omega'$ where $\omega' = \sqrt{\omega_0^2 - \beta^2}$. Hence

$$\boxed{x(t) = a_0\,e^{-\beta t}\cos(\omega' t + \phi)}$$

This represents oscillation at the reduced angular frequency $\omega'$ with an amplitude $a_0 e^{-\beta t}$ that decays exponentially.

Significance of the damping constant $\beta$:

• It fixes the rate of decay of amplitude; the amplitude falls to $1/e$ of its value in time $\tau = 1/\beta$ (relaxation time).
• It lowers the oscillation frequency from $\omega_0$ to $\omega' = \sqrt{\omega_0^2-\beta^2}$.
• Its size relative to $\omega_0$ decides whether the motion is underdamped ($\beta<\omega_0$), critically damped ($\beta=\omega_0$), or overdamped ($\beta>\omega_0$).

(b) Quality factor of a tuning fork [5 marks]

Given $f = 256$ Hz, $Q = 1000$.

The energy decays as $E = E_0 e^{-2\beta t} = E_0 e^{-t/\tau_E}$, where the energy-decay time is $\tau_E = 1/(2\beta)$.

The quality factor is $Q = \omega_0 \tau_E = \omega_0/(2\beta)$, with $\omega_0 = 2\pi f$.

Time for energy to fall to $1/e$:

$$\tau_E = \frac{Q}{\omega_0} = \frac{Q}{2\pi f} = \frac{1000}{2\pi\times 256} = 0.622\ \text{s}$$

Energy lost in the first second. Fractional energy remaining after $t = 1$ s:

$$\frac{E}{E_0} = e^{-t/\tau_E} = e^{-1/0.622} = e^{-1.608} = 0.200$$

So the fraction of energy lost in the first second is

$$1 - 0.200 = 0.80 \quad\Rightarrow\quad \textbf{about 80\% of the initial energy.}$$

Results: $\tau_E \approx 0.62\ \text{s}$; energy lost in first second $\approx 80\%$ of $E_0$.

(a) Maxwell's equations and displacement current [8 marks]

Maxwell's equations in differential form:

Displacement current and continuity. Taking the divergence of the original Ampere law $\nabla\times\mathbf{H} = \mathbf{J}$ gives $\nabla\cdot\mathbf{J} = \nabla\cdot(\nabla\times\mathbf{H}) = 0$. But the equation of continuity (charge conservation) requires

$$\nabla\cdot\mathbf{J} + \frac{\partial \rho}{\partial t} = 0,$$

which is non-zero when charge density changes (e.g. a charging capacitor). Maxwell removed the contradiction by adding the displacement current density $\mathbf{J}_d = \partial\mathbf{D}/\partial t$:

$$\nabla\times\mathbf{H} = \mathbf{J} + \frac{\partial\mathbf{D}}{\partial t}.$$

Now $\nabla\cdot\left(\mathbf{J} + \dfrac{\partial\mathbf{D}}{\partial t}\right) = \nabla\cdot\mathbf{J} + \dfrac{\partial}{\partial t}(\nabla\cdot\mathbf{D}) = \nabla\cdot\mathbf{J} + \dfrac{\partial\rho}{\partial t} = 0$, exactly the continuity equation. Thus Ampere's law becomes consistent with charge conservation.

(b) Wave equation and velocity of EM waves [4 marks]

In free space $\rho=0,\ \mathbf{J}=0$, and $\mathbf{D}=\varepsilon_0\mathbf{E}$, $\mathbf{B}=\mu_0\mathbf{H}$. Take the curl of Faraday's law:

$$\nabla\times(\nabla\times\mathbf{E}) = -\frac{\partial}{\partial t}(\nabla\times\mathbf{B}) = -\mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}.$$

Using the identity $\nabla\times(\nabla\times\mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E}$ with $\nabla\cdot\mathbf{E}=0$:

$$\boxed{\nabla^2\mathbf{E} = \mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}}$$

Comparing with the standard wave equation $\nabla^2\mathbf{E} = \dfrac{1}{c^2}\dfrac{\partial^2\mathbf{E}}{\partial t^2}$ gives the velocity

$$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = \frac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}} \approx 3\times10^8\ \text{m/s},$$

the speed of light, confirming light is an electromagnetic wave.

(a) Newton's rings by reflected light [8 marks]

Arrangement. A plano-convex lens of large radius of curvature $R$ rests on a flat glass plate, enclosing a thin wedge-shaped air film of thickness increasing radially from zero at the point of contact. Monochromatic light falls normally (via a glass plate inclined at 45° above the lens) and the reflected light is viewed through a travelling microscope. Interference between light reflected from the lower surface of the lens and the top of the flat plate produces concentric bright and dark rings centred on the contact point.

Ray diagram (described): incident light → 45° glass plate reflects it down → strikes air film → partial reflection at lens–air and air–plate boundaries → recombined reflected rays travel up to microscope.

Condition. At radial distance $r$ the air-film thickness is $t$. From the geometry of the sphere, $r^2 = (2R-t)t \approx 2Rt$, so $t = r^2/(2R)$.

For reflected light, with an extra path difference of $\lambda/2$ due to reflection at the denser plate surface (air film, $\mu=1$, normal incidence):

• Dark rings (destructive): $2t = n\lambda$, giving $r_n^2 = nR\lambda$, so

$$\boxed{r_n^{(\text{dark})} = \sqrt{nR\lambda}} \quad (n=0,1,2,\dots)$$

The radii are proportional to $\sqrt{n}$.

• Bright rings (constructive): $2t = (2n-1)\lambda/2$, giving

$$\boxed{r_n^{(\text{bright})} = \sqrt{\frac{(2n-1)R\lambda}{2}}}$$

so radii are proportional to $\sqrt{2n-1}$.

Why the centre is dark. At the point of contact $t=0$, so the geometrical path difference is zero, but the $\lambda/2$ phase change on reflection at the glass plate makes the net path difference $\lambda/2$ — the condition for destructive interference. Hence the central spot is dark.

(b) Wavelength calculation [4 marks]

For dark rings, $D_n^2 = 4nR\lambda$. Using two rings:

$$D_{m}^2 - D_{n}^2 = 4(m-n)R\lambda \quad\Rightarrow\quad \lambda = \frac{D_{m}^2 - D_{n}^2}{4(m-n)R}.$$

Given $D_{10}=0.50$ cm, $D_{20}=0.70$ cm, $m-n = 10$, $R = 100$ cm:

$$\lambda = \frac{(0.70)^2 - (0.50)^2}{4(10)(100)} = \frac{0.49 - 0.25}{4000} = \frac{0.24}{4000}\ \text{cm}$$ $$\lambda = 6.0\times10^{-5}\ \text{cm} = 6000\ \text{\AA} = 600\ \text{nm}.$$

(a) Time-independent Schrodinger equation (1-D) [7 marks]

A free or bound particle is described by a wave function $\Psi(x,t)$. For a particle of mass $m$ and total energy $E$ in a potential $V(x)$, write the time-dependent equation

$$i\hbar\frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V(x)\Psi.$$

Separation of variables. For a stationary state, $\Psi(x,t) = \psi(x)\,e^{-iEt/\hbar}$. Substituting and cancelling the time factor $e^{-iEt/\hbar}$:

$$E\psi = -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi.$$

Rearranging gives the time-independent Schrodinger equation:

$$\boxed{\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}\big(E - V\big)\psi = 0}$$

(Alternative derivation from $E=p^2/2m+V$ with operator substitutions $E\to i\hbar\partial_t$, $p\to -i\hbar\partial_x$ is equally valid.)

Conditions for an acceptable wave function $\psi$:

1. $\psi$ must be single-valued at every point.
2. $\psi$ and its first derivative $d\psi/dx$ must be continuous and finite everywhere.
3. $\psi$ must be normalizable: $\int_{-\infty}^{\infty}|\psi|^2\,dx = 1$ (so $\psi\to0$ at infinity).
4. $\psi$ must be well-behaved (no discontinuities), so that $|\psi|^2$ is a valid probability density.

(b) Particle in a 1-D infinite well of width $L$ [5 marks]

Inside the well $0<x<L$, $V=0$; outside, $V=\infty$ so $\psi=0$. Inside:

$$\frac{d^2\psi}{dx^2} + k^2\psi = 0,\qquad k^2 = \frac{2mE}{\hbar^2}.$$

General solution $\psi = A\sin kx + B\cos kx$. Boundary conditions: $\psi(0)=0 \Rightarrow B=0$; $\psi(L)=0 \Rightarrow \sin kL = 0 \Rightarrow kL = n\pi$ ($n=1,2,3,\dots$). Hence $k = n\pi/L$.

Energy eigenvalues:

$$\boxed{E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}}\qquad n=1,2,3,\dots$$

Energy is quantized; the lowest (zero-point) energy is $E_1 = h^2/8mL^2 \neq 0$.

Normalization: $\displaystyle\int_0^L A^2\sin^2\!\left(\frac{n\pi x}{L}\right)dx = A^2\frac{L}{2} = 1 \Rightarrow A = \sqrt{2/L}$.

$$\boxed{\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right),\quad 0<x<L}$$

Resonance in a forced mechanical oscillator [8 marks]

A damped oscillator driven by a periodic force $F_0\cos pt$ obeys

$$m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = F_0\cos pt,$$

or, with $2\beta=b/m$, $\omega_0^2=k/m$, $f_0=F_0/m$:

$$\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = f_0\cos pt.$$

Steady-state solution. Try $x = a\cos(pt-\theta)$. Substituting and equating coefficients gives the amplitude

$$\boxed{a = \frac{f_0}{\sqrt{(\omega_0^2 - p^2)^2 + 4\beta^2 p^2}}}, \qquad \tan\theta = \frac{2\beta p}{\omega_0^2 - p^2}.$$

Resonance (amplitude maximum). $a$ is maximum when the denominator is minimum, i.e. when

$$\frac{d}{dp}\Big[(\omega_0^2-p^2)^2 + 4\beta^2 p^2\Big] = 0 \Rightarrow -4p(\omega_0^2-p^2) + 8\beta^2 p = 0,$$

giving the resonant (driving) frequency

$$p_r = \sqrt{\omega_0^2 - 2\beta^2}.$$

At this frequency the maximum amplitude is

$$a_{max} = \frac{f_0}{2\beta\sqrt{\omega_0^2 - \beta^2}}.$$

For light damping ($\beta \to 0$), $p_r \to \omega_0$ and $a_{max}\to f_0/(2\beta\omega_0)$ becomes very large — sharp resonance.

Resonance curves (described): plotting amplitude $a$ versus driving frequency $p$:

• For small damping the curve is tall and sharply peaked near $p=\omega_0$.
• As damping increases, the peak becomes lower, broader, and shifts slightly to lower frequency.
• For heavy damping the peak disappears, amplitude falling monotonically with $p$.

Thus resonance is the condition of maximum response of the oscillator to the driving force, sharpest for low damping.

(a) States of polarization [3 marks]

Light is a transverse wave; the state of polarization describes how the electric-field vector $\mathbf{E}$ behaves at a point.

• Plane (linearly) polarized: $\mathbf{E}$ oscillates along a single fixed direction; the tip of $\mathbf{E}$ traces a straight line. Arises when two perpendicular components are in phase (or one is absent).
• Circularly polarized: two perpendicular components of equal amplitude with a phase difference of $\pi/2$; the tip of $\mathbf{E}$ traces a circle, rotating with constant magnitude.
• Elliptically polarized: two perpendicular components of unequal amplitude (or phase difference other than $0,\pi,\pi/2$); the tip of $\mathbf{E}$ traces an ellipse. It is the general case, of which linear and circular are special cases.

(b) Nicol prism [5 marks]

Construction. A Nicol prism is made from a rhombohedral crystal of calcite (Iceland spar). The crystal is cut along its diagonal and the two halves are cemented together with Canada balsam, whose refractive index (1.55) lies between the refractive indices of calcite for the ordinary ray ($\mu_o = 1.658$) and the extraordinary ray ($\mu_e = 1.486$).

Working. When unpolarized light enters, double refraction splits it into:

• the ordinary ray (O-ray), for which balsam is optically rarer ($\mu_o > \mu_{balsam}$); at the balsam layer it strikes beyond the critical angle and is totally internally reflected to the blackened side, where it is absorbed.
• the extraordinary ray (E-ray), for which balsam is optically denser ($\mu_e < \mu_{balsam}$); it is transmitted with little deviation.

The emergent beam is therefore plane-polarized (only the E-ray), vibrating in the principal plane.

As polarizer and analyzer: A first Nicol used to produce plane-polarized light from unpolarized light acts as the polarizer. A second Nicol placed in the path examines that light as the analyzer: rotating it varies the transmitted intensity per Malus's law $I = I_0\cos^2\theta$, with zero intensity (crossed Nicols) when its principal plane is perpendicular to that of the polarizer.

(a) Plane diffraction grating and grating equation [5 marks]

A plane diffraction grating is an optical component consisting of a large number of equally spaced, parallel, fine slits (rulings) ruled on a transparent (or reflecting) plate. If $a$ is the slit width and $b$ the opaque spacing, the grating element is $d = a+b$; for $N$ lines per unit length, $d = 1/N$.

Grating equation. When a plane wavefront falls normally on the grating, light diffracted at angle $\theta$ from successive slits has a path difference $d\sin\theta$ between adjacent slits. For all slits to interfere constructively (principal maxima), this path difference must be an integral multiple of $\lambda$:

$$\boxed{d\sin\theta = n\lambda}, \qquad n = 0,1,2,\dots$$

where $n$ is the order of the maximum. This is the grating equation giving the angular positions of the principal maxima. The central ($n=0$) maximum is undeviated; higher orders are progressively more diffracted, and the grating separates wavelengths since $\theta$ depends on $\lambda$.

(b) Angle of second-order diffraction [3 marks]

Given $N = 5000$ lines/cm, so

$$d = \frac{1}{N} = \frac{1}{5000}\ \text{cm} = 2\times10^{-4}\ \text{cm} = 2\times10^{-6}\ \text{m}.$$

Wavelength $\lambda = 5890\ \text{\AA} = 5.89\times10^{-7}\ \text{m}$, order $n=2$.

$$\sin\theta = \frac{n\lambda}{d} = \frac{2\times 5.89\times10^{-7}}{2\times10^{-6}} = \frac{1.178\times10^{-6}}{2\times10^{-6}} = 0.589.$$ $$\theta = \sin^{-1}(0.589) \approx 36.1^\circ.$$

(a) Emission processes and population inversion [5 marks]

• Spontaneous emission: an atom in an excited state $E_2$ decays on its own to a lower state $E_1$, emitting a photon of energy $h\nu = E_2-E_1$. The emitted photons are random in phase, direction and polarization (incoherent), as in ordinary light sources.
• Stimulated emission: an incident photon of energy $h\nu = E_2-E_1$ induces an excited atom to drop to $E_1$, emitting a second photon identical in frequency, phase, direction and polarization. The light is thus amplified and coherent.
• Population inversion: a non-equilibrium condition in which the number of atoms in the higher energy state exceeds that in the lower state ($N_2 > N_1$), opposite to the normal Boltzmann distribution.

Why population inversion is essential: In equilibrium $N_1 > N_2$, so absorption dominates over stimulated emission and a light beam is attenuated. Net amplification (laser action) requires stimulated emission to exceed absorption, which is only possible when $N_2 > N_1$. Hence population inversion (achieved by pumping) is a necessary condition for sustained laser action.

(b) He-Ne laser [3 marks]

Construction. A gas discharge tube contains a mixture of helium and neon (roughly 10:1) at low pressure, with mirrors (one fully reflecting, one partly reflecting) forming the optical cavity. A high voltage produces an electric discharge.

Working (four-level scheme, described energy diagram):

1. Electrons in the discharge collide with He atoms, exciting them to metastable levels $2^1S$ and $2^3S$ (about 20.61 eV and 19.81 eV).
2. These energies nearly coincide with neon's $5s$ and $4s$ levels, so excited He atoms transfer energy to Ne atoms by resonant collisions, creating population inversion in the Ne upper levels.
3. Lasing transition: Ne atoms drop from the $5s$/$4s$ levels to $3p$ levels giving the familiar red output at 632.8 nm (also IR lines).
4. Ne atoms then quickly fall from $3p$ to $3s$ and back to ground (the $3s\to$ ground de-excitation occurs through wall collisions), maintaining the inversion.

The coherent 632.8 nm beam is amplified by repeated reflection in the cavity and emerges through the partial mirror.

(a) Acceptance angle and numerical aperture [5 marks]

Acceptance angle ($\theta_a$): the maximum angle, measured from the fibre axis, at which light entering the core is guided by total internal reflection at the core–cladding interface (rays incident at larger angles escape into the cladding).

Numerical aperture (NA): the light-gathering ability of the fibre, defined as $NA = \sin\theta_a$ (in air).

Derivation. Let the core have refractive index $n_1$, cladding $n_2$ ($n_1 > n_2$), surrounding medium $n_0$ (air, $n_0=1$). For total internal reflection at the core–cladding boundary the angle there must equal or exceed the critical angle $\theta_c$, where $\sin\theta_c = n_2/n_1$.

A ray entering the end face at angle $\theta_a$ refracts into the core at angle $\phi$. By Snell's law at the entry face:

$$n_0\sin\theta_a = n_1\sin\phi.$$

The ray then strikes the core–cladding wall at angle $(90^\circ-\phi)$, and the limiting condition is $(90^\circ-\phi)=\theta_c$, so $\phi = 90^\circ-\theta_c$ and $\sin\phi = \cos\theta_c$.

$$\therefore\ n_0\sin\theta_a = n_1\cos\theta_c = n_1\sqrt{1-\sin^2\theta_c} = n_1\sqrt{1-\frac{n_2^2}{n_1^2}} = \sqrt{n_1^2 - n_2^2}.$$

With $n_0=1$:

$$\boxed{NA = \sin\theta_a = \sqrt{n_1^2 - n_2^2}}$$

(b) Numerical [3 marks]

Given $n_1 = 1.50$, $n_2 = 1.47$.

$$NA = \sqrt{n_1^2 - n_2^2} = \sqrt{1.50^2 - 1.47^2} = \sqrt{2.25 - 2.1609} = \sqrt{0.0891} = 0.298.$$ $$\theta_a = \sin^{-1}(0.298) \approx 17.4^\circ.$$

Results: Numerical aperture $\approx 0.30$, acceptance angle $\approx 17.4^\circ$.

(a) Polarization, displacement and the relation D = ε₀E + P [5 marks]

Electric polarization ($\mathbf{P}$): the electric dipole moment induced per unit volume of a dielectric when placed in an electric field. Its magnitude equals the surface density of bound (polarization) charge.

Electric displacement ($\mathbf{D}$): a vector field defined so that its flux depends only on free charge; $\oint \mathbf{D}\cdot d\mathbf{A} = Q_{free}$.

Relation. Inside a dielectric the total (bound + free) charge produces $\mathbf{E}$. Gauss's law in vacuum gives $\varepsilon_0\nabla\cdot\mathbf{E} = \rho_{total} = \rho_{free} + \rho_{bound}$, with $\rho_{bound} = -\nabla\cdot\mathbf{P}$. Therefore

$$\nabla\cdot(\varepsilon_0\mathbf{E} + \mathbf{P}) = \rho_{free}.$$

Defining the displacement as the quantity whose divergence is the free charge:

$$\boxed{\mathbf{D} = \varepsilon_0\mathbf{E} + \mathbf{P}}$$

Dielectric constant and susceptibility. For a linear dielectric, $\mathbf{P} = \varepsilon_0\chi_e\mathbf{E}$, where $\chi_e$ is the electric susceptibility. Then

$$\mathbf{D} = \varepsilon_0\mathbf{E} + \varepsilon_0\chi_e\mathbf{E} = \varepsilon_0(1+\chi_e)\mathbf{E} = \varepsilon_0\varepsilon_r\mathbf{E}.$$ $$\boxed{\varepsilon_r = 1 + \chi_e}$$

The dielectric constant (relative permittivity) exceeds unity by the susceptibility.

(b) Capacitance [3 marks]

Given $A = 100\ \text{cm}^2 = 100\times10^{-4} = 10^{-2}\ \text{m}^2$, $d = 1\ \text{mm} = 10^{-3}\ \text{m}$, $\varepsilon_r = 5$.

$$C = \frac{\varepsilon_0\varepsilon_r A}{d} = \frac{(8.85\times10^{-12})(5)(10^{-2})}{10^{-3}}.$$ $$C = \frac{8.85\times10^{-12}\times5\times10^{-2}}{10^{-3}} = 4.425\times10^{-10}\ \text{F} \approx 443\ \text{pF}.$$

(a) Heisenberg's uncertainty principle [5 marks]

Statement. It is impossible to determine simultaneously and with arbitrary precision both the position and the momentum of a particle. The product of the uncertainties is at least of order $\hbar$:

$$\Delta x\,\Delta p_x \ge \frac{\hbar}{2}, \qquad \hbar = \frac{h}{2\pi}.$$

It follows from the wave nature of matter: a localized particle is a wave packet built from a spread of momenta, so narrowing $\Delta x$ widens $\Delta p_x$. (Similar relations hold for energy–time, $\Delta E\,\Delta t \ge \hbar/2$.)

Electron cannot exist inside the nucleus. A nucleus has radius $\sim 10^{-14}$ m, so if an electron were confined inside, $\Delta x \approx 10^{-14}$ m. Then

$$\Delta p \ge \frac{\hbar}{2\Delta x} = \frac{1.05\times10^{-34}}{2\times10^{-14}} \approx 5.3\times10^{-21}\ \text{kg\,m/s}.$$

The minimum momentum $p \gtrsim 5.3\times10^{-21}$ kg·m/s is so large that the electron is relativistic; using $E \approx pc$:

$$E \approx pc = 5.3\times10^{-21}\times3\times10^8 \approx 1.6\times10^{-12}\ \text{J} \approx 10\ \text{MeV}.$$

No electron emitted from nuclei (beta particles) is observed with energies of this order (they are typically a few MeV or less, and binding energies cannot hold such an electron). Hence an electron of such enormous energy cannot remain bound inside the nucleus — electrons do not exist within the nucleus.

(b) de Broglie wavelength of an electron [3 marks]

For an electron accelerated through potential difference $V$, $eV = \dfrac{p^2}{2m}$, so $p = \sqrt{2meV}$ and

$$\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}\ \text{\AA} \quad (V\ \text{in volts}).$$

For $V = 100$ V:

$$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227\ \text{\AA} = 1.23\times10^{-10}\ \text{m}.$$

(a) Hall effect [5 marks]

Definition. When a current-carrying conductor or semiconductor is placed in a magnetic field perpendicular to the current, a transverse voltage (Hall voltage) develops across the specimen, perpendicular to both the current and the field. This is the Hall effect.

Derivation. Let current $I$ flow along $x$ in a slab of thickness $t$ and width $w$, with magnetic field $B$ along $z$. Carriers (charge $q$, drift velocity $v_d$, density $n$) experience the magnetic force $qv_dB$, deflecting them sideways until the built-up transverse electric field $E_H$ balances it:

$$qE_H = qv_dB \Rightarrow E_H = v_dB.$$

The Hall voltage is $V_H = E_H w = v_d B w$. Since $I = nqv_d(wt)$, $v_d = \dfrac{I}{nq\,wt}$, giving

$$V_H = \frac{IB}{nqt}.$$

The Hall coefficient is defined as $R_H = \dfrac{E_H}{J_x B} = \dfrac{1}{nq}$, so

$$\boxed{R_H = \frac{1}{nq} = \frac{V_H t}{IB}}.$$

Use in semiconductors:

• The sign of $R_H$ (or $V_H$) gives the carrier type: positive for p-type (holes), negative for n-type (electrons).
• The magnitude gives the carrier concentration $n = 1/(|R_H|q)$.
• Combined with conductivity $\sigma$, the mobility follows from $\mu = |R_H|\sigma$.

(b) Type I vs Type II superconductors; Meissner effect [3 marks]

Meissner effect: When a material is cooled below its critical temperature in a magnetic field, it expels the magnetic flux from its interior completely ($\mathbf{B}=0$ inside), behaving as a perfect diamagnet. This shows superconductivity is not merely perfect conductivity but a distinct thermodynamic state.