Q.A resistor $R = 6\ \Omega$ is connected across a battery of emf $V = 6\ \text{V}$ of negligible internal resistance, forming a single loop in which a steady current $I$ flows.

The problem gives you a fixed potential difference across a wire — 240 V at all temperatures. As the wire heats up, its resistance increases, which means the current must drop. But we are asked for the current at the cold end, $0^\circ\mathrm{C}$.

The central relationship here is how resistance changes with temperature for a metallic conductor. Over a moderate range, it is linear:

where $R_0$ is the resistance at $0^\circ\mathrm{C}$, $\alpha$ is the temperature coefficient of resistance, and $T$ is the temperature in $^\circ\mathrm{C}$.

We know the resistance increase when going from $0^\circ\mathrm{C}$ to $1000^\circ\mathrm{C}$ is exactly 25 $\Omega$. That piece of information, together with $\alpha$, lets us solve for $R_0$. Once we have $R_0$, the current at $0^\circ\mathrm{C}$ is simply $I = V / R_0$.

Let’s work it through.

1. Write the change in resistance. At $T = 1000^\circ\mathrm{C}$, the resistance is $R_{1000} = R_0 (1 + \alpha \cdot 1000)$. The increase is:

The problem states this increase equals 25 $\Omega$. So:

2. Solve for $R_0$. $1.25 \times 10^{-4} \times 1000 = 0.125$. So: $R_0 \times 0.125 = 25$ …

The key idea is that the wires are made of the same metal, so their resistivity $\rho$ is identical. They are joined end to end, forming a series circuit. The free ends are held at fixed potentials — 20 V on the longer wire’s free end, 2 V on the shorter wire’s free end. This means a steady current flows from the higher potential to the lower one, and the junction potential is somewhere between them.

Since the wires are in series, the same current $I$ flows through both. The potential difference across each wire is $V = IR$, where $R$ depends on the wire’s geometry. The junction potential is simply the potential at the free end of the shorter wire plus the voltage drop across that wire (or equivalently, the potential at the free end of the longer wire minus the drop across it).

Let’s work it out step by step.

1. Find the resistance of each wire. For a cylindrical wire, $R = \rho \frac{L}{A}$, where $A = \pi r^2$. Let the shorter wire (length $L_1 = 100$ cm, radius $r_1 = 0.6$ mm) have resistance $R_1$, and the longer wire ($L_2 = 180$ cm, $r_2 = 0.9$ mm) have resistance $R_2$. Since $\rho$ and $\pi$ are common, we can work with ratios:

Compute:

Simplify: $\frac{100}{180} = \frac{5}{9}$, and $\frac{0.9}{0.6} = 1.5$, so $(1.5)^2 = 2.25 = \frac{9}{4}$.

Thus:

So $R_1 : R_2 = 5 : 4$. The shorter wire has the larger resistance.

2. Set up the series circuit. The total potential difference across the combination is $20\,\text{V} - 2\,\text{V} = 18\,\text{V}$. In a series circuit, the voltage drop across each resistor is proportional to its resistance:

Using the ratio $R_1 : R_2 = 5 : 4$, we have:

Therefore:

$V_2 = \frac{4}{9} \times 18 = 8\,\text{V}.$ …

The key idea is that in a doped semiconductor, the product of electron and hole concentrations remains constant at a given temperature, equal to the square of the intrinsic carrier concentration. This is the law of mass action for semiconductors.

When we add arsenic (a group V element) to silicon, each arsenic atom donates one extra electron to the crystal. The silicon already has $6 \times 10^{28}$ atoms per cubic metre, and we add 1 part per million (ppm) of arsenic. That means for every million silicon atoms, there is one arsenic atom. So the donor concentration $N_D$ is:

At room temperature, nearly all these donor atoms are ionised, so the electron concentration $n_e$ in the n-type silicon is essentially equal to $N_D$ (since the intrinsic contribution is negligible compared to this huge doping). So:

Now, the intrinsic carrier concentration $n_i$ is given as $1.2 \times 10^{16} \ \mathrm{m}^{-3}$. The mass-action law states:

Therefore, the hole concentration $n_h$ is:

Let's compute that step by step:

1. Square the intrinsic concentration: $(1.2 \times 10^{16})^2 = 1.44 \times 10^{32}$ …

The key idea is that the root-mean-square speed of gas molecules is given by

$v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$, where $T$ is the absolute temperature (in Kelvin).

For a fixed gas (same molecules), $m$ and $k$ are constant, so $v_{\text{rms}} \propto \sqrt{T}$.

Thus, if we know the speed at one temperature, we can find it at another by taking the ratio of the square roots of the absolute temperatures.

1. Convert temperatures to Kelvin

   The Celsius-to-Kelvin conversion is $T(\mathrm{K}) = T(^\circ\mathrm{C}) + 273$.

   • At $77^\circ\mathrm{C}$: $T_1 = 77 + 273 = 350\ \mathrm{K}$
   • At $150.5^\circ\mathrm{C}$: $T_2 = 150.5 + 273 = 423.5\ \mathrm{K}$

2. Write the proportionality

   Since $v_{\text{rms}} \propto \sqrt{T}$, we have

where $v_1 = 50\ \mathrm{ms}^{-1}$, $T_1 = 350\ \mathrm{K}$, $T_2 = 423.5\ \mathrm{K}$.

3. Compute the ratio

(Notice $423.5 = 350 \times 1.21$ exactly, since $350 \times 1.21 = 423.5$.)

4. Take the square root

(Because $1.1^2 = 1.21$.)

5. Find the new speed $$$$

The problem is about a doped semiconductor. In an intrinsic (pure) semiconductor, the number of electrons equals the number of holes, both equal to the intrinsic carrier concentration $n_i$. When we dope it with an impurity — here, clearly a donor impurity since the electron concentration increases dramatically — the balance shifts. But a fundamental rule holds at any given temperature: the product of electron and hole concentrations stays constant, equal to $n_i^2$. This is the law of mass action.

Let’s work through it.

1. Write down what is given. Intrinsic carrier concentration: $n_i = 1.2 \times 10^{16} \ \mathrm{m}^{-3}$. After doping, the electron concentration $n_e$ becomes $20 \times 10^5$ times the initial (intrinsic) concentration. So

2. Simplify $n_e$.

3. Apply the law of mass action. For any semiconductor at equilibrium,

where $n_h$ is the hole concentration.

So

4. Substitute the numbers.

First, square the numerator:

Then divide: …

Concept & Intuition

In a parallel plate capacitor, the displacement current $I_d$ through the dielectric is exactly equal to the conduction current $I$ in the connecting wires. This is a cornerstone of Maxwell’s equations: a changing electric field between the plates produces a displacement current that “completes” the circuit. For a capacitor, the conduction current is $I = C \frac{dV}{dt}$, where $\frac{dV}{dt}$ is the rate at which the voltage across the plates changes. So if we know the displacement current and the capacitance, we can directly find $\frac{dV}{dt}$.

Step-by-step solution

1. Recall the relation between displacement current and voltage change For a parallel plate capacitor, the displacement current $I_d$ is given by

where $\Phi_E$ is the electric flux. But since $Q = CV$ and the electric field is uniform, this simplifies to

This is the key formula: the displacement current equals the capacitance times the rate of change of voltage.

2. Identify the given values

   • Capacitance: $C = 30\,\mu\mathrm{F} = 30 \times 10^{-6}\,\mathrm{F}$
   • Displacement current: $I_d = 150\,\mu\mathrm{A} = 150 \times 10^{-6}\,\mathrm{A}$

3. Solve for $\frac{dV}{dt}$

   From $I_d = C \frac{dV}{dt}$, we rearrange:

   $$$$

Why electromagnetic waves slow down in matter

In vacuum, electromagnetic waves travel at the universal constant $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$. When light enters a material medium, it interacts with the atoms: the oscillating electric field polarizes the material (captured by permittivity $\varepsilon$) and the oscillating magnetic field magnetizes it (captured by permeability $\mu$). These interactions cause the wave to propagate more slowly.

The speed in any medium is given by

where $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$. Since $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$, we can write

The product $\mu_r \varepsilon_r$ tells us how much the medium "resists" the wave compared to vacuum. The refractive index $n = \sqrt{\mu_r \varepsilon_r}$ for non-magnetic materials simplifies to $\sqrt{\varepsilon_r}$, but here both parameters matter.

Calculation

1. Identify the given parameters:

   • Relative permeability: $\mu_r = 2.3$
   • Relative permittivity: $\varepsilon_r = 1.3$
   • Speed of light in vacuum: $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$

2. Compute the product under the square root: …

Because B is at zero potential, the potential at any other point is just the voltage change across the resistor joining it to B, and the currents are fixed by Kirchhoff's current law (KCL) at the junction.

Step 1 — KCL at B.

The algebraic sum of the currents meeting at junction B is zero. Writing the branch currents through the resistors at B and applying KCL fixes each branch current consistently with $V_B = 0$.

Step 2 — Potential at A. …

Concept & Intuition

A current-carrying wire produces a magnetic field that circles around it. An electron moving parallel to the wire cuts across these field lines, so it experiences a magnetic force. The key is to first find the magnetic field at the electron’s location, then apply the Lorentz force law. Since the electron’s velocity is parallel to the wire, the force is perpendicular to both the velocity and the field — and we can compute its magnitude directly.

Step-by-step solution

1. Magnetic field due to the wire For a long straight wire, the magnetic field at a perpendicular distance $r$ is

Here $\mu_0 = 4\pi \times 10^{-7}$ SI, $I = 2.0$ A, and $r = 1$ cm $= 0.01$ m.

Substitute:

2. Force on the moving electron The magnetic force on a charge $q$ moving with velocity $v$ in a field $B$ is $F = qvB \sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. The electron moves parallel to the wire. The magnetic field lines are circles around the wire, so at the electron’s position, $\vec{B}$ is perpendicular to the radial line — and thus perpendicular to the wire. Hence $\vec{v}$ is perpendicular to $\vec{B}$ ($\theta = 90^\circ$), so $\sin\theta = 1$. Therefore: $F = |q| v B.$ …

Concept & Intuition

Arsenic is a pentavalent donor: each atom contributes one extra electron to the conduction band. At room temperature, almost all donor atoms are ionized, so the electron concentration $n$ is approximately equal to the donor concentration $N_D$ (since $N_D \gg n_i$). The hole concentration $p$ is then determined by the law of mass action: $n p = n_i^2$, where $n_i$ is the intrinsic carrier concentration. The ratio $n/p$ is therefore $n^2 / n_i^2 \approx N_D^2 / n_i^2$.

Step‑by‑step reasoning

1. Identify the dominant carrier Arsenic is a donor impurity. Doping with $4.5 \times 10^{21}$ atoms/m³ gives an electron concentration

(The intrinsic concentration $n_i = 1.5 \times 10^{16} \ \text{m}^{-3}$ is negligible compared to $N_D$, so thermal generation adds almost nothing.)

2. Apply mass‑action law In thermal equilibrium, the product of electron and hole concentrations is constant:

Thus the hole concentration is

3. Compute the ratio $n/p$

Substitute the numbers: …