Important Derivations

Electric field at P due to $+q$ (along the direction away from $+q$, i.e., towards the right):

Electric field at P due to $-q$ (towards $-q$, i.e., towards the left):

Since $E_1 > E_2$ (as P is closer to $+q$), the net field at P is directed along the dipole moment (towards the right):

Simplify the expression inside the brackets:

For a short dipole, $r \gg a$, so $a^2 \ll r^2$. Therefore $(r^2 - a^2)^2 \approx r^4$. Also, $2qa = p$ (dipole moment):

The magnitudes of electric fields at P due to $+q$ and $-q$ are equal (since both charges are equidistant from P):

The fields $E_1$ and $E_2$ are directed away from $+q$ and towards $-q$ respectively. Resolve each into components along the axial and equatorial directions. The components perpendicular to the dipole axis (vertical components) cancel out by symmetry.

The components parallel to the dipole axis add up. Each contributes $E_1 \cos\theta$, where $\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}$. The net field is directed opposite to $\vec{p}$:

Substitute $p = q \cdot 2a$:

For a short dipole $(r \gg a)$, we have $(r^2 + a^2)^{3/2} \approx r^3$:

Force on charge $+q$: $\vec{F_1} = q\vec{E}$ (along $\vec{E}$). Force on charge $-q$: $\vec{F_2} = -q\vec{E}$ (opposite to $\vec{E}$).

Net force = $\vec{F_1} + \vec{F_2} = q\vec{E} - q\vec{E} = 0$. So the dipole does not undergo translational motion, but experiences a torque (couple).

The magnitude of the torque = force $\times$ perpendicular distance between the two forces. The perpendicular distance is $2a\sin\theta$:

In vector form, using the cross product:

Gauss’s Law states: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$

By symmetry, the electric field $\vec{E}$ is perpendicular to the sheet and points away from it on both sides (for positive $\sigma$). The magnitude is the same on both sides at equal distances.

For the cylindrical Gaussian surface, the flux through the curved surface is zero (since $\vec{E} \perp d\vec{A}$ on the curved surface, making $\vec{E} \cdot d\vec{A} = 0$).

The total flux is only through the two flat faces:

The charge enclosed by the Gaussian surface is $q_{\text{enclosed}} = \sigma A$. Applying Gauss’s Law:

By Biot-Savart law, the magnetic field due to a small element $d\vec{l}$ at point P is:

Since $d\vec{l} \perp \vec{r}$ always (the element is tangent to the loop and $\vec{r}$ points from the element to P), we have $|d\vec{l} \times \hat{r}| = dl$.

The magnitude of the field due to each element is:

Resolve $d\vec{B}$ into components: one along the axis and one perpendicular to it. By symmetry, the perpendicular components from diametrically opposite elements cancel out. Only the axial components survive.

Integrate over the entire loop ($\oint dl = 2\pi R$):

Simplify:

Special case: At the centre of the loop ($x = 0$): $B = \frac{\mu_0 I}{2R}$

The magnetic field produced by conductor 1 at the location of conductor 2 (at distance $d$) is given by Ampere’s circuital law:

Conductor 2, carrying current $I_2$, is placed in this field $B_1$. The force on a length $L$ of conductor 2 is:

Force per unit length:

Nature of force: Using Fleming’s left-hand rule, if currents are in the same direction, the force is attractive. If currents are in opposite directions, the force is repulsive.

Definition of Ampere: When $I_1 = I_2 = 1\text{ A}$ and $d = 1\text{ m}$, the force per unit length is $F/L = 2 \times 10^{-7}\text{ N/m}$.

The magnetic field inside a long solenoid is uniform and given by:

The magnetic flux through each turn of the solenoid is:

The total flux linkage with the solenoid (all $N$ turns) is:

By definition, self-inductance $L$ is given by $N\phi = LI$. Therefore:

(since $N = nl$, so $N^2/l = n^2 l$)

Let the current in the circuit be $I = I_0 \sin(\omega t)$ (same current flows through all elements in series).

The voltage across each element:

• Across $R$: $V_R = IR$ (in phase with $I$)
• Across $L$: $V_L = IX_L = I\omega L$ (leads $I$ by $90°$)
• Across $C$: $V_C = IX_C = \frac{I}{\omega C}$ (lags $I$ by $90°$)

In the phasor diagram, $V_L$ and $V_C$ are opposite in direction. Their resultant is $(V_L - V_C)$ directed upward (assuming $V_L > V_C$). This resultant is perpendicular to $V_R$.

By Pythagoras’ theorem, the amplitude of the resultant voltage is:

Factor out $I$:

Since $V = IZ$ (where $Z$ is the impedance), we get:

The phase angle $\phi$ between voltage and current is given by:

If $X_L > X_C$: circuit is inductive, voltage leads current. If $X_L < X_C$: circuit is capacitive, current leads voltage.

From the impedance expression derived above:

The current amplitude is $I_0 = V_0 / Z$. Current is maximum when impedance $Z$ is minimum.

$Z$ is minimum when the reactive part vanishes, i.e., when:

Solving for the resonant angular frequency $\omega_0$:

The resonant frequency in terms of $f$:

Properties at resonance:

• Impedance is minimum: $Z_{\min} = R$
• Current is maximum: $I_{\max} = V_0 / R$
• Phase angle $\phi = 0$ — voltage and current are in phase
• $V_L = V_C$ (they cancel each other)
• The circuit behaves as a purely resistive circuit
• Power factor $\cos\phi = 1$ (maximum power is dissipated)

When an alternating voltage $V_p$ is applied to the primary coil, it produces an alternating current which creates a time-varying magnetic flux $\phi$ in the core.

Since the core channels all the flux, the same flux $\phi$ links both coils. By Faraday’s law of electromagnetic induction, the induced EMF in the primary is:

Similarly, the induced EMF in the secondary is:

Dividing the two equations (and noting that for an ideal transformer, $\varepsilon_p \approx V_p$ and $\varepsilon_s \approx V_s$):

If $k > 1$: step-up transformer ($V_s > V_p$). If $k < 1$: step-down transformer ($V_s < V_p$).

For an ideal transformer, there are no energy losses. By conservation of energy, input power = output power:

Therefore, the current ratio is:

A step-up transformer increases voltage but decreases current proportionally, and vice versa.

Energy losses in a real transformer:

• Copper losses: $I^2R$ heating in the windings (reduced by using thick copper wire)
• Eddy current losses: Induced currents in the core (reduced by using laminated core)
• Hysteresis loss: Energy spent in repeated magnetisation (reduced by using soft iron core)
• Flux leakage: Not all flux links both coils (reduced by winding coils over each other)

For a ray from O incident at point M on the surface, let the angles made with the principal axis by OM, MI, and MC be $\alpha$, $\beta$, and $\gamma$ respectively. The angle of incidence is $i$ and angle of refraction is $r$.

From the geometry of the triangle OMC: the exterior angle $i = \alpha + \gamma$.

From triangle MCI: the exterior angle $\gamma = r + \beta$, so $r = \gamma - \beta$.

For small angles (paraxial ray approximation): $\sin i \approx i$ and $\sin r \approx r$. Applying Snell’s law: $n_1 \sin i = n_2 \sin r$:

Substituting: $n_1(\alpha + \gamma) = n_2(\gamma - \beta)$

For small angles, using $\tan\theta \approx \theta$ and the perpendicular MN from M to the axis:

(Here $u$ is negative as the object is on the left of P, by sign convention: $OP = -u$.)

Substituting into $n_1(\alpha + \gamma) = n_2(\gamma - \beta)$:

Cancel $MN$ throughout:

Rearranging:

Refraction at surface $S_1$: Light goes from medium of refractive index $n_1$ to $n_2$. Object distance is $u$ and the intermediate image is at distance $v'$. Using the result for refraction at a single spherical surface:

Refraction at surface $S_2$: Light goes from medium $n_2$ to $n_1$. The intermediate image $I'$ at distance $v'$ from $S_1$ acts as the object for $S_2$. For a thin lens, we take the object distance for $S_2$ as $v'$ itself. The final image is at distance $v$:

Add equations (i) and (ii). The $n_2/v'$ terms cancel:

Divide both sides by $n_1$:

When the object is at infinity ($u \to \infty$), the image is formed at the focal point ($v = f$). Setting $\frac{1}{u} = 0$ and writing $\mu = n_2/n_1$:

Let point P on the screen be at distance $y$ from the central point O (the perpendicular bisector of $S_1S_2$).

The distances from $S_1$ and $S_2$ to P are:

Path difference: $\Delta = S_2P - S_1P$. We compute $S_2P^2 - S_1P^2$:

Also, $S_2P^2 - S_1P^2 = (S_2P - S_1P)(S_2P + S_1P)$. Since $D \gg d$ and $D \gg y$, we have $S_1P \approx S_2P \approx D$, so $S_2P + S_1P \approx 2D$:

Condition for bright fringes (constructive interference): path difference = $n\lambda$

Condition for dark fringes (destructive interference): path difference = $(2n-1)\frac{\lambda}{2}$

Fringe width: The distance between two consecutive bright (or dark) fringes:

The fringe width is the same for both bright and dark fringes and is independent of $n$ (i.e., all fringes are equally spaced).

A photon of frequency $\nu$ has energy:

When this photon strikes the metal surface, part of its energy is used to overcome the work function $W_0$ (the minimum energy needed to release the electron from the metal):

where $\nu_0$ is the threshold frequency (minimum frequency below which no photoelectric emission occurs).

The remaining energy appears as kinetic energy of the emitted electron. By conservation of energy:

where $KE_{\max} = \frac{1}{2}mv_{\max}^2$ is the maximum kinetic energy of the emitted photoelectron.

Since $KE_{\max} = eV_0$ (where $V_0$ is the stopping potential), we can also write:

Or equivalently: $eV_0 = h\nu - W_0 = h(\nu - \nu_0)$.

For an electron in a circular orbit, the centripetal force is provided by the Coulomb electrostatic force:

From Bohr’s quantisation condition (Postulate 2):

Deriving the radius: Substitute equation (ii) into equation (i):

Simplifying:

Solving for $r$:

For $n = 1$ (ground state, Bohr radius): $r_1 = a_0 = 0.529$ Å $\approx 0.53 \times 10^{-10}$ m. In general, $r_n = n^2 a_0$.

Deriving the velocity: Substituting $r_n$ back into equation (ii):

Note that $v_n \propto 1/n$ — the electron moves slower in higher orbits.

Deriving the energy: The total energy of the electron is the sum of kinetic and potential energies.

Kinetic energy: From equation (i), $KE = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r}$

Potential energy: $PE = -\frac{e^2}{4\pi\varepsilon_0 r}$

Substituting $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$:

Substituting numerical values: