Class 12 Physics Chapter-Wise Subjective Question Bank PDF — 15 Questions Per Chapter (SA & LA with Solutions)

(a) Ohm's Law: It states that the current $I$ flowing through a conductor is directly proportional to the potential difference $V$ across its ends, provided the physical conditions (like temperature) remain unchanged. $V \propto I$ or $V = IR$.
Resistance ($R$): It is the opposition to the flow of current. It is defined as the ratio of the potential difference across the conductor to the current flowing through it, $R = V/I$. The SI unit of resistance is the ohm ($\Omega$).

(b) Derivation: We know the drift velocity of electrons is given by $v_d = \frac{eE}{m}\tau$, where $E$ is the electric field, $\tau$ is the relaxation time, $e$ is the charge of an electron, and $m$ is its mass.
The current density $J$ is related to drift velocity by $J = n e v_d$, where $n$ is the number density of free electrons.
Substituting the value of $v_d$, we get:
$J = n e (\frac{eE}{m}\tau) = \frac{ne^2\tau}{m} E$
Since for a given conductor, $n$, $e$, $m$, and $\tau$ are constants, we can write $\frac{ne^2\tau}{m} = \sigma$, where $\sigma$ is the conductivity of the material.
Therefore, $J = \sigma E$. This is the vector form of Ohm's law.

(a) Kirchhoff's Rules:
1. Junction Rule (First Law): The algebraic sum of currents entering a junction in an electrical circuit is zero. $\sum I = 0$. This means that the total current entering a junction must equal the total current leaving it.
2. Loop Rule (Second Law): The algebraic sum of the changes in potential around any closed loop involving resistors and cells in a circuit is zero. $\sum \Delta V = 0$.

(b) Explanation with diagram:
[A simple circuit diagram showing a junction (e.g., three wires meeting) and a closed loop with at least one cell and two resistors should be drawn here.]
For the junction rule: At junction P, if currents $I_1$ and $I_2$ are entering and $I_3$ is leaving, then according to the rule, $I_1 + I_2 = I_3$ or $I_1 + I_2 - I_3 = 0$.
For the loop rule: In a closed loop (e.g., ABCDA), we sum up the potential drops (across resistors, in the direction of current) and potential gains (across cells, from negative to positive terminal). The total sum is zero. For example, in a loop with a cell of emf $\epsilon$ and resistors $R_1$ and $R_2$ with current $I$, we can write $\epsilon - IR_1 - IR_2 = 0$.

Conservation Principles:
- Kirchhoff's junction rule is based on the law of conservation of charge.
- Kirchhoff's loop rule is based on the law of conservation of energy.

(a) Principle: The Wheatstone bridge works on the principle of null deflection, i.e., the ratio of two resistances in one arm is equal to the ratio of the other two resistances in the other arm when no current flows through the galvanometer.

(b) Circuit Diagram:
[A standard Wheatstone bridge diagram should be drawn, showing four resistors $P, Q, R, S$ in a quadrilateral arrangement. A cell is connected across one diagonal (e.g., AC) and a galvanometer across the other diagonal (e.g., BD).]

(c) Derivation of Balance Condition:
Let the current from the cell be $I$. At junction A, it splits into $I_1$ (through arm ADB) and $I_2$ (through arm ACB).
In the balanced state, the current through the galvanometer is zero ($I_g = 0$). This means the potential at point B is equal to the potential at point D, i.e., $V_B = V_D$.
Applying Kirchhoff's loop rule to loop ABDA:
$I_1 P - I_2 R = 0$ (since $I_g=0$)
$\Rightarrow I_1 P = I_2 R$ ... (1)
Applying Kirchhoff's loop rule to loop BCDB:
$I_1 Q - I_2 S = 0$ (since $I_g=0$)
$\Rightarrow I_1 Q = I_2 S$ ... (2)
Dividing equation (1) by equation (2), we get:
$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S}$
$\Rightarrow \frac{P}{Q} = \frac{R}{S}$
This is the condition for the balanced state of the Wheatstone bridge.

(a) Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the physical conditions (like temperature) remain unchanged. Mathematically, $V \propto I$ or $V = IR$, where R is the constant of proportionality called resistance. The SI unit of resistance is Ohm ($\Omega$). One Ohm is the resistance of a conductor through which a current of one ampere flows when a potential difference of one volt is applied across its ends.
(b) The graph of $I$ versus $V$ is a straight line passing through the origin. The resistance ($R$) can be found from the reciprocal of the slope of the $I-V$ graph. $Slope = \frac{\Delta I}{\Delta V} = \frac{1}{R}$. Therefore, $R = \frac{1}{Slope}$.
(c) Given $V = 20$ V, $R = 5 \Omega$. Using Ohm's law, $I = \frac{V}{R} = \frac{20}{5} = 4$ A.

(a) Kirchhoff's Laws:
1. Junction Rule (First Law): The algebraic sum of currents entering a junction in an electrical circuit is zero. $\sum I = 0$.
2. Loop Rule (Second Law): The algebraic sum of the changes in potential around any closed loop involving resistors and cells is zero. $\sum \Delta V = 0$.

(b) Basis of the laws:
1. Junction Rule is based on the law of conservation of charge. It implies that charge cannot accumulate or be drained from a junction.
2. Loop Rule is based on the law of conservation of energy. It means that the net change in potential energy of a charge after traversing a closed loop is zero.

(c) According to Kirchhoff's junction rule, the total current entering the junction must equal the total current leaving the junction.
Total incoming current = 2 A + 3 A + 4 A = 9 A.
Total outgoing current = $I$.
Therefore, $I = 9$ A.

a) An equipotential surface is a surface with a constant value of potential at all points on the surface.
Two properties of equipotential surfaces are:
1. No work is done in moving a test charge from one point to another on an equipotential surface.
2. The electric field is always perpendicular to the equipotential surface at every point.

b) (i) For a single point charge ($q > 0$), the equipotential surfaces are concentric spheres centered at the charge.
(ii) For a uniform electric field, the equipotential surfaces are planes normal to the electric field lines.

(a) Let $q$ be the charge on the capacitor at some instant during charging and $v$ be the potential difference across it. We have $v = q/C$. The work done to add a small charge $dq$ is $dW = v dq = \frac{q}{C} dq$.
The total work done in charging the capacitor from $0$ to $Q$ is stored as potential energy $U$.
$U = W = \int_{0}^{Q} \frac{q}{C} dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_0^Q = \frac{Q^2}{2C}$.
Since $Q=CV$, we can also write $U = \frac{(CV)^2}{2C} = \frac{1}{2}CV^2$ or $U = \frac{1}{2}QV$.

(b)
(i) For parallel combination, the equivalent capacitance is $C_p = C_1 + C_2 + C_3$.
$C_p = 2 \text{ pF} + 3 \text{ pF} + 4 \text{ pF} = 9 \text{ pF}$.

(ii) In a parallel combination, the voltage across each capacitor is the same, $V = 100 \text{ V}$.
Charge on the first capacitor, $Q_1 = C_1 V = 2 \times 10^{-12} \text{ F} \times 100 \text{ V} = 200 \times 10^{-12} \text{ C} = 200 \text{ pC}$.
Charge on the second capacitor, $Q_2 = C_2 V = 3 \times 10^{-12} \text{ F} \times 100 \text{ V} = 300 \times 10^{-12} \text{ C} = 300 \text{ pC}$.
Charge on the third capacitor, $Q_3 = C_3 V = 4 \times 10^{-12} \text{ F} \times 100 \text{ V} = 400 \times 10^{-12} \text{ C} = 400 \text{ pC}$.

(a) Electrostatic potential at any point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that point without any acceleration.
S.I. unit: Volt (V) or Joules/Coulomb ($J C^{-1}$).

(b) Consider a point charge $+Q$ at the origin O. Let P be a point at a distance r from O. To find the potential at P, we calculate the work done in bringing a unit positive charge from infinity to P.
Let the unit positive charge be at point A, at a distance $x$ from O. The electrostatic force on it is $F = \frac{1}{4\pi\epsilon_0} \frac{Q \cdot 1}{x^2}$.
The work done to move the charge by a small distance $dx$ against this force is $dW = -F dx = -\frac{1}{4\pi\epsilon_0} \frac{Q}{x^2} dx$.
The total work done in bringing the charge from infinity to point P is:
$W = \int_{\infty}^{r} dW = \int_{\infty}^{r} -\frac{1}{4\pi\epsilon_0} \frac{Q}{x^2} dx$
$W = -\frac{Q}{4\pi\epsilon_0} \int_{\infty}^{r} x^{-2} dx = -\frac{Q}{4\pi\epsilon_0} [-\frac{1}{x}]_{\infty}^{r}$
$W = \frac{Q}{4\pi\epsilon_0} [\frac{1}{r} - \frac{1}{\infty}] = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.
By definition, this work done is the potential at point P. So, $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.

(a) Capacitance of a conductor is defined as the ratio of the charge given to the conductor to the potential raised on it. $C = Q/V$. It depends on the size and shape of the conductor, and the nature of the surrounding medium.

(b) The electric field in the air gap between the plates is $E_0 = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$.
The electric field inside the dielectric slab is $E = \frac{E_0}{K} = \frac{Q}{AK\epsilon_0}$.
The potential difference between the plates is the sum of potential differences across the air gap (distance $d-t$) and the dielectric slab (distance $t$).
$V = E_0(d-t) + E(t)$
$V = \frac{Q}{A\epsilon_0}(d-t) + \frac{Q}{AK\epsilon_0}(t)$
$V = \frac{Q}{A\epsilon_0} \left( (d-t) + \frac{t}{K} \right)$
Capacitance $C = \frac{Q}{V}$.
$C = \frac{Q}{\frac{Q}{A\epsilon_0} \left( (d-t) + \frac{t}{K} \right)}$
$C = \frac{A\epsilon_0}{(d-t) + \frac{t}{K}}$.

(a) Derivation for potential at a general point P(r, θ) due to a dipole:
Let the dipole have charges $-q$ and $+q$ separated by distance $2a$. The potential at P is $V = V_1 + V_2 = \frac{1}{4\pi\epsilon_0} [\frac{q}{r_1} - \frac{q}{r_2}]$.
For a short dipole ($r >> a$), we find $r_1 \approx r - a \cos\theta$ and $r_2 \approx r + a \cos\theta$.
Substituting and simplifying, we get $V = \frac{q}{4\pi\epsilon_0} [\frac{1}{r - a \cos\theta} - \frac{1}{r + a \cos\theta}] = \frac{q}{4\pi\epsilon_0} \frac{2a \cos\theta}{r^2 - a^2 \cos^2\theta}$.
Since $r >> a$, $a^2 \cos^2\theta$ is negligible. Using $p = q(2a)$, we get $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$.
Axial point: $\theta = 0^\circ$, so $V_{axial} = \frac{p}{4\pi\epsilon_0 r^2}$.
Equatorial point: $\theta = 90^\circ$, so $V_{equatorial} = 0$.

(b) Given: $q = 1 \mu C = 1 \times 10^{-6} C$, $2a = 2 \text{ cm} = 0.02 \text{ m}$, $r = 10 \text{ cm} = 0.1 \text{ m}$.
Dipole moment, $p = q(2a) = (1 \times 10^{-6})(0.02) = 2 \times 10^{-8} \text{ C m}$.
For an axial point, potential $V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2 - a^2}$.
Here $a=0.01$ m. $V = (9 \times 10^9) \times \frac{2 \times 10^{-8}}{(0.1)^2 - (0.01)^2} = \frac{180}{0.01 - 0.0001} = \frac{180}{0.0099} \approx 1.82 \times 10^4 \text{ V}$.

(a) Ohm's Law: It states that the current $I$ flowing through a conductor is directly proportional to the potential difference $V$ across its ends, provided the physical conditions (like temperature) remain unchanged. $V \propto I$ or $V = IR$.
Resistance ($R$): It is the opposition to the flow of current. It is defined as the ratio of the potential difference across the conductor to the current flowing through it, $R = V/I$. The SI unit of resistance is the ohm ($\Omega$).

(b) Derivation: We know the drift velocity of electrons is given by $v_d = \frac{eE}{m}\tau$, where $E$ is the electric field, $\tau$ is the relaxation time, $e$ is the charge of an electron, and $m$ is its mass.
The current density $J$ is related to drift velocity by $J = n e v_d$, where $n$ is the number density of free electrons.
Substituting the value of $v_d$, we get:
$J = n e (\frac{eE}{m}\tau) = \frac{ne^2\tau}{m} E$
Since for a given conductor, $n$, $e$, $m$, and $\tau$ are constants, we can write $\frac{ne^2\tau}{m} = \sigma$, where $\sigma$ is the conductivity of the material.
Therefore, $J = \sigma E$. This is the vector form of Ohm's law.

(a) Kirchhoff's Rules:
1. Junction Rule (First Law): The algebraic sum of currents entering a junction in an electrical circuit is zero. $\sum I = 0$. This means that the total current entering a junction must equal the total current leaving it.
2. Loop Rule (Second Law): The algebraic sum of the changes in potential around any closed loop involving resistors and cells in a circuit is zero. $\sum \Delta V = 0$.

(b) Explanation with diagram:
[A simple circuit diagram showing a junction (e.g., three wires meeting) and a closed loop with at least one cell and two resistors should be drawn here.]
For the junction rule: At junction P, if currents $I_1$ and $I_2$ are entering and $I_3$ is leaving, then according to the rule, $I_1 + I_2 = I_3$ or $I_1 + I_2 - I_3 = 0$.
For the loop rule: In a closed loop (e.g., ABCDA), we sum up the potential drops (across resistors, in the direction of current) and potential gains (across cells, from negative to positive terminal). The total sum is zero. For example, in a loop with a cell of emf $\epsilon$ and resistors $R_1$ and $R_2$ with current $I$, we can write $\epsilon - IR_1 - IR_2 = 0$.

Conservation Principles:
- Kirchhoff's junction rule is based on the law of conservation of charge.
- Kirchhoff's loop rule is based on the law of conservation of energy.

(a) Principle: The Wheatstone bridge works on the principle of null deflection, i.e., the ratio of two resistances in one arm is equal to the ratio of the other two resistances in the other arm when no current flows through the galvanometer.

(b) Circuit Diagram:
[A standard Wheatstone bridge diagram should be drawn, showing four resistors $P, Q, R, S$ in a quadrilateral arrangement. A cell is connected across one diagonal (e.g., AC) and a galvanometer across the other diagonal (e.g., BD).]

(c) Derivation of Balance Condition:
Let the current from the cell be $I$. At junction A, it splits into $I_1$ (through arm ADB) and $I_2$ (through arm ACB).
In the balanced state, the current through the galvanometer is zero ($I_g = 0$). This means the potential at point B is equal to the potential at point D, i.e., $V_B = V_D$.
Applying Kirchhoff's loop rule to loop ABDA:
$I_1 P - I_2 R = 0$ (since $I_g=0$)
$\Rightarrow I_1 P = I_2 R$ ... (1)
Applying Kirchhoff's loop rule to loop BCDB:
$I_1 Q - I_2 S = 0$ (since $I_g=0$)
$\Rightarrow I_1 Q = I_2 S$ ... (2)
Dividing equation (1) by equation (2), we get:
$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S}$
$\Rightarrow \frac{P}{Q} = \frac{R}{S}$
This is the condition for the balanced state of the Wheatstone bridge.

(a) Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the physical conditions (like temperature) remain unchanged. Mathematically, $V \propto I$ or $V = IR$, where R is the constant of proportionality called resistance. The SI unit of resistance is Ohm ($\Omega$). One Ohm is the resistance of a conductor through which a current of one ampere flows when a potential difference of one volt is applied across its ends.
(b) The graph of $I$ versus $V$ is a straight line passing through the origin. The resistance ($R$) can be found from the reciprocal of the slope of the $I-V$ graph. $Slope = \frac{\Delta I}{\Delta V} = \frac{1}{R}$. Therefore, $R = \frac{1}{Slope}$.
(c) Given $V = 20$ V, $R = 5 \Omega$. Using Ohm's law, $I = \frac{V}{R} = \frac{20}{5} = 4$ A.

(a) Kirchhoff's Laws:
1. Junction Rule (First Law): The algebraic sum of currents entering a junction in an electrical circuit is zero. $\sum I = 0$.
2. Loop Rule (Second Law): The algebraic sum of the changes in potential around any closed loop involving resistors and cells is zero. $\sum \Delta V = 0$.

(b) Basis of the laws:
1. Junction Rule is based on the law of conservation of charge. It implies that charge cannot accumulate or be drained from a junction.
2. Loop Rule is based on the law of conservation of energy. It means that the net change in potential energy of a charge after traversing a closed loop is zero.

(c) According to Kirchhoff's junction rule, the total current entering the junction must equal the total current leaving the junction.
Total incoming current = 2 A + 3 A + 4 A = 9 A.
Total outgoing current = $I$.
Therefore, $I = 9$ A.

a) Ampere's Circuital Law: It states that the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the total current $I_{enc}$ enclosed by the loop.
Mathematically: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$

b) Magnetic field inside a solenoid:
Consider a long solenoid of length $L$ with $N$ turns, so the number of turns per unit length is $n = N/L$. Let $I$ be the current flowing through it.
To find the magnetic field inside, we consider a rectangular Amperian loop $abcd$ as shown in the diagram, with length $h$.
The line integral of $\vec{B}$ over the closed loop $abcd$ is:
$\oint_{abcd} \vec{B} \cdot d\vec{l} = \int_{a}^{b} \vec{B} \cdot d\vec{l} + \int_{b}^{c} \vec{B} \cdot d\vec{l} + \int_{c}^{d} \vec{B} \cdot d\vec{l} + \int_{d}^{a} \vec{B} \cdot d\vec{l}$
For an ideal solenoid, the magnetic field outside is zero, so $\int_{c}^{d} \vec{B} \cdot d\vec{l} = 0$.
Also, $\vec{B}$ is perpendicular to $d\vec{l}$ along paths $bc$ and $da$, so $\int_{b}^{c} \vec{B} \cdot d\vec{l} = 0$ and $\int_{d}^{a} \vec{B} \cdot d\vec{l} = 0$.
Therefore, the integral is non-zero only along the path $ab$ inside the solenoid.
$\oint \vec{B} \cdot d\vec{l} = \int_{a}^{b} B dl \cos(0^\circ) = B \int_{a}^{b} dl = Bh$.
The total current enclosed by the loop is the number of turns within the loop multiplied by the current $I$. Number of turns in length $h$ is $nh$. So, $I_{enc} = nhI$.
Applying Ampere's law: $Bh = \mu_0 (nhI)$.
Thus, the magnetic field inside the solenoid is $B = \mu_0 n I$.

a) Force on a current-carrying conductor: The force $\vec{F}$ experienced by a straight conductor of length vector $\vec{l}$ carrying current $I$ in a uniform magnetic field $\vec{B}$ is given by:
$\vec{F} = I (\vec{l} \times \vec{B})$
The magnitude of the force is $F = I l B \sin\theta$, where $\theta$ is the angle between the conductor and the magnetic field.
The force is maximum when $\sin\theta = 1$, which means $\theta = 90^\circ$. This occurs when the conductor is placed perpendicular to the direction of the magnetic field.

b) Moving Coil Galvanometer:
Principle: It is based on the principle that a current-carrying loop placed in a uniform magnetic field experiences a torque.
Working: A rectangular coil is suspended between the pole pieces of a strong permanent magnet. When a current $I$ flows through the coil, a torque acts on it. The torque is given by $\tau = NIAB \sin\phi$, where $N$ is the number of turns, $A$ is the area of the coil, $B$ is the magnetic field, and $\phi$ is the angle between the normal to the coil and the magnetic field. The magnetic field is made radial, so $\phi = 90^\circ$ always, and the torque is $\tau = NIAB$. This torque deflects the coil. A restoring torque is produced in the suspension spring, given by $\tau_{restore} = k\alpha$, where $k$ is the torsional constant of the spring and $\alpha$ is the angular deflection. In equilibrium, the deflecting torque equals the restoring torque.
$NIAB = k\alpha$
Therefore, the deflection is directly proportional to the current: $\alpha = (\frac{NAB}{k})I$. The deflection is measured by a pointer attached to the coil.

a) Biot-Savart Law: According to this law, the magnetic field $d\vec{B}$ at a point P due to a small current element $I d\vec{l}$ is given by:
$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3}$
where $\mu_0$ is the permeability of free space, $I$ is the current, $d\vec{l}$ is the length element of the conductor, and $\vec{r}$ is the position vector from the element to the point P.

b) Derivation for a circular loop:
Consider a circular loop of radius $R$ carrying a current $I$. We want to find the magnetic field at a point P on its axis at a distance $x$ from the center.
Consider a small element $d\vec{l}$ on the loop. The distance of P from this element is $r = \sqrt{R^2 + x^2}$.
The magnetic field due to this element at P is $dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(90^\circ)}{r^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2}$.
The direction of this field is perpendicular to the plane containing $d\vec{l}$ and $\vec{r}$.
We resolve $d\vec{B}$ into two components: $dB_x = dB \cos\theta$ along the axis, and $dB_\perp$ perpendicular to the axis.
Due to symmetry, the perpendicular components cancel out. The net field is the sum of the axial components.
$B = \int dB_x = \int dB \cos\theta$.
From the diagram, $\cos\theta = \frac{R}{r} = \frac{R}{\sqrt{R^2 + x^2}}$.
$B = \int \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2} \cdot \frac{R}{\sqrt{R^2 + x^2}} = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \int dl$.
Since $\int dl = 2\pi R$ (circumference of the loop),
$B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} (2\pi R) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
For a coil with $N$ turns, the field is $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$.

a) **Biot-Savart's Law:** It states that the magnitude of the magnetic field doldsymbol{B} due to a current element doldsymbol{l} carrying a current $I$ at a point with position vector oldsymbol{r} from the element is directly proportional to the current $I$, the length of the element doldsymbol{l}, the sine of the angle $\theta$ between doldsymbol{l} and oldsymbol{r}, and inversely proportional to the square of the distance $r$.
Mathematically, $dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$.
**Vector Form:** $d\boldsymbol{B} = \frac{\mu_0}{4\pi} \frac{I (d\boldsymbol{l} \times \boldsymbol{r})}{r^3}$.

b) **Derivation for a circular loop:**
Consider a circular loop of radius $R$ carrying current $I$. We want to find the magnetic field at its centre $O$.
Consider a small current element doldsymbol{l} on the loop. The position vector oldsymbol{R} from the element to the centre is the radius vector.
The angle between doldsymbol{l} and oldsymbol{R} is always $90^\circ$ (i.e., $\theta = 90^\circ$).
According to Biot-Savart's law, the magnetic field at the centre due to this element is:
$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(90^\circ)}{R^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2}$
The direction of this field is perpendicular to the plane of the loop, given by the right-hand thumb rule.
To find the total magnetic field $B$, we integrate $dB$ over the entire length of the loop (circumference $2\pi R$):
$B = \int dB = \int_0^{2\pi R} \frac{\mu_0 I}{4\pi R^2} dl$
$B = \frac{\mu_0 I}{4\pi R^2} \int_0^{2\pi R} dl = \frac{\mu_0 I}{4\pi R^2} [l]_0^{2\pi R}$
$B = \frac{\mu_0 I}{4\pi R^2} (2\pi R)$
$B = \frac{\mu_0 I}{2R}$
For a coil with $N$ turns, $B = \frac{\mu_0 N I}{2R}$.

a) **Ampere's Circuital Law:** It states that the line integral of the magnetic field $\boldsymbol{B}$ around any closed loop (Amperian loop) is equal to $\mu_0$ times the total current $I_{enc}$ passing through the surface enclosed by the loop.
Mathematically, $\oint \boldsymbol{B} \cdot d\boldsymbol{l} = \mu_0 I_{enc}$.

b) **Magnetic Field inside a Solenoid:**
Consider a long solenoid with 'n' turns per unit length carrying a current $I$. The magnetic field inside a long solenoid is uniform and parallel to its axis, while it is nearly zero outside.
To find the field inside, we consider a rectangular Amperian loop $pqrs$ as shown in a diagram. Let the length of the side $pq$ be $L$.
The line integral of $\boldsymbol{B}$ over the closed loop $pqrs$ is:
$\oint_{pqrs} \boldsymbol{B} \cdot d\boldsymbol{l} = \int_{p}^{q} \boldsymbol{B} \cdot d\boldsymbol{l} + \int_{q}^{r} \boldsymbol{B} \cdot d\boldsymbol{l} + \int_{r}^{s} \boldsymbol{B} \cdot d\boldsymbol{l} + \int_{s}^{p} \boldsymbol{B} \cdot d\boldsymbol{l}$
- For path $pq$ (inside), $\boldsymbol{B}$ is parallel to $d\boldsymbol{l}$, so $\theta=0^\circ$. $\int_{p}^{q} \boldsymbol{B} \cdot d\boldsymbol{l} = \int BL dl = BL$.
- For paths $qr$ and $sp$, $\boldsymbol{B}$ is perpendicular to $d\boldsymbol{l}$, so $\theta=90^\circ$, and the integral is zero.
- For path $rs$ (outside), the magnetic field $\boldsymbol{B}$ is approximately zero, so the integral is zero.
Thus, $\oint \boldsymbol{B} \cdot d\boldsymbol{l} = BL$.
Total number of turns enclosed by the loop is $nL$. The total current enclosed is $I_{enc} = nLI$.
Applying Ampere's law, $\oint \boldsymbol{B} \cdot d\boldsymbol{l} = \mu_0 I_{enc}$.
$BL = \mu_0 (nLI)$
$B = \mu_0 n I$.

(a) Definitions:
(i) Magnetic Declination: The angle between the geographic meridian and the magnetic meridian at a place is called the magnetic declination at that place.
(ii) Angle of Dip: The angle that the total magnetic field $B_E$ of the Earth makes with the surface of the Earth (horizontal direction) is the angle of dip.
(iii) Horizontal Component: It is the component of the total magnetic field of the Earth in the horizontal direction. It is given by $B_H = B_E · Γ¥σ δ$.

(b) Given:
Declination, $D = 12^°$ West
Angle of Dip, $δ = 60^°$
Horizontal component, $B_H = 0.16$ G

We know that $B_H = B_E · Γ¥σ δ$.
So, the magnitude of the Earth's magnetic field is:
$B_E = Β_H / Γ¥σ δ = 0.16 / Γ¥σ 60^° = 0.16 / 0.5 = 0.32$ G.
The direction is $12^°$ west of geographic north, and the field is directed at an angle of $60^°$ upwards from the horizontal.

(a) Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. Mathematically, $Φ_B = ∮ Β · dΑ = 0$.
Significance: This law implies that magnetic monopoles (isolated north or south poles) do not exist. Magnetic poles always exist in pairs (dipoles).

(b) The magnetic field lines for a bar magnet are shown originating from the North pole and terminating at the South pole outside the magnet, and from South to North inside the magnet, forming closed loops.
Properties of magnetic field lines:
1. Magnetic field lines are continuous closed loops.
2. The tangent to the field line at any point gives the direction of the net magnetic field at that point.
3. The density of field lines in a region represents the strength of the magnetic field. They are crowded in regions of strong field and are far apart in regions of weak field.
4. Two magnetic field lines never intersect each other.

The differences between Dia-, Para-, and Ferro-magnetic substances are as follows:

| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| 1. Behaviour in field | Feebly repelled by magnets. | Feebly attracted by magnets. | Strongly attracted by magnets. |
| 2. Susceptibility ($χ$) | Small and negative (e.g., $χ ≈ -10^{-5}$) | Small and positive (e.g., $χ ≈ 10^{-5}$) | Large and positive (e.g., $χ ≈ 10^3$) |
| 3. Permeability ($μ_r$) | Slightly less than 1 ($μ_r < 1$) | Slightly greater than 1 ($μ_r > 1$) | Much greater than 1 ($μ_r » 1$) |
| 4. Temperature effect | Susceptibility is independent of temperature. | Obeys Curie's Law, $χ ∝ 1/T$. | Above Curie temperature, it becomes paramagnetic. |
| 5. Example | Bismuth, Copper, Water | Aluminium, Sodium, Oxygen | Iron, Cobalt, Nickel |

(a) A magnetic field line is a curve, the tangent to which at any point gives the direction of the magnetic field at that point.

Properties of magnetic field lines:
1. They form continuous closed loops.
2. The tangent to the field line at any point gives the direction of the magnetic field at that point.
3. The larger the density of field lines, the stronger the magnetic field.
4. They do not intersect each other. If they did, it would mean there are two directions of the magnetic field at the point of intersection, which is not possible.

(b) The magnetic field lines of a bar magnet are shown in a diagram. They emerge from the North pole and enter the South pole outside the magnet. Inside the magnet, they travel from the South pole to the North pole, forming closed loops.

(a) Magnetic Declination ($\\delta$): It is the angle between the geographic meridian and the magnetic meridian at a place. It represents the angle by which a compass needle is deflected from the true north.

(b) Angle of Dip or Inclination ($I$): It is the angle that the Earth's total magnetic field ($B_E$) makes with the horizontal direction in the magnetic meridian. At the magnetic equator, the dip is zero, and at the magnetic poles, it is 90^\\circ.

(c) Horizontal component of Earth's magnetic field ($B_H$): It is the component of the Earth's total magnetic field ($B_E$) in the horizontal direction in the magnetic meridian. It is given by $B_H = B_E \\cos(I)$. It is this component that is used by a compass to find direction.

(a) Electromagnetic waves are waves that consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation. They do not require a material medium to travel.
They are produced by accelerated electric charges. For example, an oscillating charge produces an oscillating electric field, which in turn produces an oscillating magnetic field, and this process continues, generating an EM wave.

(b) Four properties of electromagnetic waves are:
1. They are transverse in nature.
2. They travel with the speed of light in vacuum, $c = 3 \times 10^8$ m/s.
3. They do not require any material medium for their propagation.
4. The electric field vector ($\\vec{E}$) and magnetic field vector ($\\vec{B}$) are mutually perpendicular to each other and also to the direction of propagation of the wave.