Solution: The energy density of an electromagnetic wave is equal for the electric and magnetic fields:
U_E = (1/2)ε₀E² and U_B = (1/2μ₀)B²
Since E = cB, both U_E and U_B are equal.
Final Conclusion: Electric and magnetic fields in an electromagnetic wave carry equal energy densities.

Quick Tip: In electromagnetic waves, energy is shared equally between the electric and magnetic fields.

Solution:
- Impedance depends on both reactance and resistance.
- Capacitive reactance decreases with frequency (X_c = 1/ωC).
- Inductive reactance increases with frequency (X_L = ωL).
- Resistance is frequency independent.
Final Conclusion: Match physical properties with their dependence on frequency.

Quick Tip: For circuits, remember the frequency dependencies of reactance for capacitors and inductors.

Solution: The correct order of electromagnetic waves in decreasing wavelength is:
Radio waves > Microwaves > Infrared > Visible > X-rays
Final Conclusion: Electromagnetic spectrum is ordered by wavelength from radio to gamma rays.

Quick Tip: Remember: Longer wavelength corresponds to lower frequency and energy.

Solution:
- Microwaves are produced by magnetrons.
- Infrared waves arise from vibrations of atoms/molecules.
- X-rays are generated by bombarding heavy nuclei with electrons.
- Radio waves are created by LC oscillators.
Final Conclusion: Match each electromagnetic wave with its production method accurately.

Quick Tip: Understand the specific production methods for different electromagnetic waves.

Solution: Using the refraction formula for curved surfaces:
^μ₂⁄_v - ^μ₁⁄_u = ^{μ₂ - μ₁}⁄_R
Substitute:
¹⁄_v - ^4⁄₃⁄_-20 = ^{1 - 4⁄₃}⁄₁₀
Simplify to get v = -16 cm, meaning the image is 16 cm to the left of P in air.
Final Conclusion: The position of the image is determined by the refraction formula for spherical surfaces.

Quick Tip: Sign conventions in optics are crucial. Remember to assign positive or negative signs based on direction.

Solution: Lens power is given by:
P = (μ - 1)(¹⁄_R1 - ¹⁄_R2)
For fixed radii, power is directly proportional to (μ - 1).
Final Conclusion: Lens power depends on refractive index and curvature.

Quick Tip: For lenses, changes in refractive index significantly affect the focal length and power.

Solution: The lens equation is:
¹⁄_f = ¹⁄_v - ¹⁄_u
This gives a hyperbolic relation between v and u, with asymptotes along the axes.
Final Conclusion: A convex lens produces a hyperbolic relation between v and u.

Quick Tip: Analyze lens behavior by rearranging the lens equation and plotting for various u values.

Solution: The linear width of the central maximum is:
Δx = ^2λL⁄_a
Substitute Δx = 5mm, L = 50cm, and a = 0.1mm:
λ = ^{Δx * a}⁄_2L
λ = ^{(5 * 10-3)(0.1 * 10-3)}⁄_(2)(0.5) = 5 x 10^-7m
Final Conclusion: The central maximum width helps determine wavelength in single-slit diffraction.

Quick Tip: Diffraction patterns depend on slit width, screen distance, and wavelength. Understand the linear width relation.

Solution: The photoelectric equation is:
K_max = hf - hv₀
Here, f = 2f₀, so:
K_max = h(2f₀ – f₀) = hf₀
Final Conclusion: Kinetic energy depends on the difference between incident and threshold frequencies.

Quick Tip: The photoelectric effect requires f > f₀, with excess energy converted into kinetic energy.

Solution: The photoelectric current depends on the intensity of light reaching the plate, which follows the inverse square law:
I ∝ ¹⁄_d²
where d is the distance of the source from the plate. However:
- For distances where the light intensity is sufficient to cause photoemission, all emitted electrons are collected if the plate is maintained at a constant potential.
- This leads to a saturation current that is independent of the distance, as long as the light intensity remains above the threshold to trigger photoemission.
Thus, the photoelectric current remains constant regardless of the distance, resulting in a horizontal line on the graph.
Conclusion: Photoelectric current reaches saturation and does not change with increasing distance if the plate's potential is adjusted to collect all emitted electrons.

Quick Tip: Photoelectric current saturates at higher intensities and remains constant as long as the intensity is above the emission threshold.

Solution: The de Broglie wavelength is given by:
λ = ^h⁄_√(2meV)
Doubling the potential V reduces λ by a factor of √2. Hence, λ decreases as V increases.
Final Conclusion: The de Broglie wavelength is inversely proportional to the square root of the accelerating potential.

Quick Tip: Increasing the kinetic energy of a particle shortens its de Broglie wavelength. For charged particles, use the relation λ ∝ ¹⁄_√K

Solution: In the Bohr model, the total energy E = -K and potential energy U = 2E = -2K. The kinetic energy is K = -E.
Final Conclusion: For electrons in the Bohr model, the total energy is negative, with potential energy twice as negative as the kinetic energy.

Quick Tip: In bound systems like atoms, kinetic energy is positive, potential energy is negative, and their sum (total energy) is negative.

Solution: The density of nuclei is constant and does not depend on their mass number, as the volume is proportional to A^1/3 and mass is proportional to A. Thus, the ratio is 1:1.
Final Conclusion: Nuclear density is independent of mass number and remains constant for all nuclei.

Quick Tip: Nuclear density is constant due to the uniform distribution of nucleons inside the nucleus.

Solution: The shortest wavelength corresponds to the transition to the lowest energy level in each series. Pfund, Brackett, Balmer, and Lyman series correspond to n = 5, 4, 2, 1, respectively, in decreasing order of wavelength.
Final Conclusion: Wavelength decreases as the final energy level (n_f) decreases.

Quick Tip: Remember the spectral series order and the fact that wavelength decreases as energy increases.

Solution: n-type semiconductors are formed by doping silicon with elements that have one more valence electron than silicon (group 15 elements). Arsenic (A) and Phosphorus (C) belong to group 15 and are suitable dopants.
Final Conclusion: To form n-type semiconductors, use dopants from group 15 with one extra electron.

Quick Tip: Group 15 elements add extra electrons, making them ideal for creating n-type semiconductors.

Solution:
Step 1: Analyze the graphs for each diode type
1. Solar cell (D): Produces current under illumination and shows current generation with almost no external voltage applied.
2. Zener diode (C): Operates in reverse bias, showing breakdown behavior at a specific voltage.
3. Forward-biased p-n junction (A): Exhibits an exponential increase in current as voltage increases in the forward direction.
4. Photodiode (B): Operates in reverse bias with a current proportional to light intensity.

Step 2: Map the sequence
- (D): Solar cell
- (C): Zener diode
- (A): Forward biased p-n junction
- (B): Photodiode
Thus, the correct sequence is: (D), (C), (A), (B)
Conclusion: Each diode's V-I characteristic corresponds to its operational mode, making the correct sequence (D), (C), (A), (B).

Quick Tip: Match V-I characteristics with the diode type based on their typical operation in forward or reverse bias.

Solution: The force on a current-carrying wire in a uniform magnetic field is given by:
**F** = I**L** x **B**
where:
- I is the current
- **L** is the length vector of the wire
- **B** is the magnetic field.

Step 1: Analyze Each Segment of the Wire
1. Straight Segment: The force on a straight segment is calculated using F = ILBsinθ, where θ = 90° (since the magnetic field is perpendicular to the plane).
2. Semicircular Segment: The magnetic force on the curved segment is equivalent to the force on its chord (net straight-line length), which is 2R.

Step 2: Combine Forces The forces from the straight segments and the curved segment contribute to a net downward force. Summing these contributions gives:
F_net = 4BIR.
Conclusion: The net force acting on the wire is 4BIR, directed vertically downward.

Quick Tip: For current-carrying wires in magnetic fields, calculate the net force by analyzing each segment individually and summing vector contributions.

Solution:
For an equilateral prism (A = 60°), the refractive index μ is related to the angle of minimum deviation (δ_m) using the formula:
μ = ^{sin((A + δ_m)/2)}⁄_sin(A/2)
Substitute μ = √2 and A = 60°:
√2 = ^{sin((60° + δ_m)/2)}⁄_sin(30°)
Simplify:
√2sin(30°) = sin^{60° + δ_m}⁄₂
^√2⁄₂ = sin^{60° + δ_m}⁄₂
This gives:
sin(45°) = sin^{60° + δ_m}⁄₂
^{60° + δ_m}⁄₂ = 45° => 60° + δ_m = 90°
Solve for δ_m:
δ_m = 30°.
Conclusion: The angle of minimum deviation is δ_m = 30°, which depends on the prism's geometry and refractive index.

Quick Tip: For prisms, use the relationship between refractive index, minimum deviation, and prism angle to determine δ_m.

Solution: Electric charge is quantized and occurs in integral multiples of the elementary charge e. This is attributed to the transfer of electrons, which carry charge e.
Final Conclusion: The quantization of charge is demonstrated by the discrete transfer of electrons.

Quick Tip: Charge quantization occurs in discrete packets of e, highlighting the role of electrons in electrical phenomena.

Solution: When the dielectric is introduced, the effective separation becomes:
d_eff = ^d-t⁄_K + t
where d = 4 mm, t = 4 mm, and d_eff becomes 4 + 3.2 = 7.2 mm. Solving for K:
7.2 = ^4-4⁄_K + 4
7.2 = ⁴⁄_K + 4 => K = 5
Final Conclusion: The dielectric constant is determined by the change in effective plate separation.

Quick Tip: For parallel plate capacitors, introducing a dielectric alters effective separation inversely proportional to K.

Solution: The upward electric force balances the downward gravitational force:
qE = V_sphere(ρ_copper - ρ_oil)g
Substitute:
q * 600π = (⁴⁄₃)π(0.5)³(8 - 0.8)10
Solving gives q = 2 x 10^-5 C.
Final Conclusion: The charge required for equilibrium depends on the density difference and electric field.

Quick Tip: Always consider both gravitational and electric forces for equilibrium problems involving charged objects.

Solution: Drift velocity is given by:
v_d = ^eEτ⁄_m
where τ (relaxation time) decreases with increasing temperature due to increased scattering, reducing v_d. Thermal velocity increases as temperature increases.
Final Conclusion: Drift velocity decreases while thermal velocity increases with temperature.

Quick Tip: Temperature impacts relaxation time and drift velocity inversely, while thermal velocity rises directly.

Solution: Simplify the combination of resistors step by step:
- Combine parallel resistors first using ¹⁄_Req = ¹⁄_R1 + ¹⁄_R2.
- Add series resistances directly. The total resistance is found to be 18 Ω.
Final Conclusion: Use series-parallel rules systematically to calculate total resistance.

Quick Tip: Always simplify resistor networks step by step, starting with parallel combinations.

Solution:
The current in the circuit is initially given by Ohm's law:
I = ^{emf of first cell}⁄_{total resistance} = ^1.1⁄_{0.5 + 0.5} = 1.1 A.
After adding the second cell in series, the total emf becomes:
Total emf = 1.1 + 1.1 = 2.2 V.
The total resistance in the circuit becomes:
R_total = 0.5 + 0.5 + r₂,
where r₂ is the internal resistance of the second cell.
For the current to remain the same (I = 1.1 A), use Ohm's law:
I = ^{Total emf}⁄_{Total resistance}.
Substitute the values:
1. 1 = ^2.2⁄_{1 + r2}
Solve for r₂:
1 + r₂ = ^2.2⁄_1.1 = 2.
r₂ = 1 Ω.
Conclusion: The internal resistance of the second cell must be 1 Ω to ensure the current remains unchanged.

Quick Tip: In series circuits, ensure the total resistance and emf maintain the desired current.

Solution: For the Wheatstone bridge to be balanced:
^P⁄_Q = ^R⁄_Sshunted
Substitute P = Q = 3 Ω, R = 3 Ω, and solve for S_shunted:
S_shunted = ^{4 x 12}⁄_{4 + 12}= 3
Since, we need the effective resistance to be 3, and S = 4. Let R be the resistance to be connected in parallel with S = 4 to get effective of 3. Then,
(1/3) = (1/4) +(1/R)
R = 12.
Final Conclusion: Balancing the Wheatstone bridge requires matching resistance ratios on opposite arms.

Quick Tip: In a Wheatstone bridge, balance the resistance ratios across the bridge arms for equilibrium.

Solution: The magnetic moment of a bar magnet is:
M = m * l
When bent into a semicircular form, the effective length becomes 2r = ^2l⁄_π The new magnetic moment is:
M' = m * ^2l⁄_π = ^2M⁄_π
Final Conclusion: Bending a magnet reduces its effective length, altering the magnetic moment.

Quick Tip: When shapes change, magnetic moment depends on the effective separation of poles.

Solution: High permeability ensures efficient magnetic flux conduction, and low hysteresis loss minimizes energy wastage during magnetization and demagnetization cycles.
Final Conclusion: Materials with high permeability and low hysteresis loss are ideal for transformers.

Quick Tip: Choose materials with low energy losses and efficient flux carrying capacity for magnetic devices.

Solution:
The induced emf in the conducting ring is given by Faraday's law:
emf = -^dΦ⁄_dt = -^{d(B * A)}⁄_dt
Substitute A = πr² (area of the ring) and ^dB⁄_dt = x (rate of change of magnetic field):
emf = -πr² * x
The induced electric field intensity E at any point on the ring is related to the emf as:
E = ^emf⁄_2πr
Substitute emf = -πr² * x:
E = ^{-πr² * x}⁄_2πr = ^rx⁄₂
Thus, the electric field intensity at any point on the ring is:
E = ^rx⁄₂
Conclusion: The electric field intensity is directly proportional to both the radius of the ring and the rate of change of the magnetic field, with a factor of ¹⁄₂ .

Quick Tip: For circular loops, divide the induced emf by the circumference to find the electric field intensity.

Solution: The induced emf in the secondary is:
e = M^dI⁄_dt
For a sinusoidal current:
I(t) = I₀sin(2πft) => ^dI⁄_dt = I₀ * 2πf
Substitute M = 0.5, I₀ = 1, and f = 50:
e= 0.5 * 1 * 2π * 50 = 100 V
Final Conclusion: Crest voltage in the secondary depends on the mutual inductance and the rate of current change.

Quick Tip: For AC transformers, calculate induced emf using the peak rate of change of current.

Solution: The emf induced in the coil is given by Faraday's law:
emf = - N ^dΦ⁄_dt, Φ = B * A
where:

• N = 100 (number of turns in the coil)
• Φ is the magnetic flux, Φ = B * A
• B = μ₀nI (magnetic field due to solenoid)
• A = πr² (area of the coil)
• n = 2 x 10⁴ turns/m (turns per meter of solenoid)
• ^dI⁄_dt= ^-4⁄_0.05 = -80 A/s
• μ₀ = 4π x 10^-7 T⋅m/A
• r = 0.01 m (radius of the coil)

Quick Tip: To solve solenoid problems, carefully compute the emf and charge using the coil's parameters and resistance.

Solution: When the lower half of a convex lens is made opaque:
- The lens still forms a complete image of the object because all parts of the object send rays through the available portion of the lens.
- Blocking part of the lens reduces the number of light rays contributing to the image. This decreases the intensity (brightness) of the image.
- The image remains complete, and its size or visibility is unaffected. Hence, the idea that "only half the object will show" (Option C) is incorrect.
Conclusion: The correct effect is described by (C): The intensity of the image gets reduced.

Quick Tip: Blocking part of a lens affects brightness but not the complete formation of the image.

Solution: The fringe separation formula is:
Δx = ^λL⁄_d
where:
- λ = 500 nm = 500 x 10^-9 m (wavelength of light)
- L = 2 m (distance to the screen)
- d = 0.1 mm = 0.1 x 10^-3 m (distance between slits)
Substitute the values into the formula:
Δx = ^{500 x 10-9 * 2}⁄_{0.1 x 10^-3}
Δx = ^{1000 x 10-9}⁄_{0.1 x 10^-3} = 0.01 m = 1 cm
Final Conclusion: Fringe separation depends linearly on wavelength and distance to the screen.

Quick Tip: Use the fringe formula Δx = ^λL⁄_d and ensure proper unit conversions.

Solution: The tube length is:
L = f_o + f_e = 10 m + 0.1 m = 1010 cm
The magnification is:
M= ^f_o⁄_fe = ^{10 m}⁄_{0.1 m} =100
Final Conclusion: Tube length and magnification depend on the focal lengths of the objective and eyepiece.

Quick Tip: For telescopes, remember L = f_o + f_e and M = ^f_o⁄_fe for direct calculations.

Solution:

• r ∝ n²: Radius depends on the square of n.
• v ∝ ¹⁄_n: Speed decreases with higher n.
• E ∝ ¹⁄_n²: Energy magnitude follows n^-2

Quick Tip: Memorize proportionality relations for Bohr's model for radius, speed, and energy.

Solution: In a full-wave rectifier, both halves of the AC cycle are used. Thus, the output frequency is twice the input frequency:
f_output = 2f_input = 2 x 50 = 100 Hz
Final Conclusion: The output frequency of a full-wave rectifier is double the input frequency.

Quick Tip: Remember, full-wave rectifiers double the input frequency by rectifying both halves of the AC cycle.

Solution: In a non-uniform field, the net force on the dipole is non-zero because the field strength varies at the two poles of the dipole. Torque is zero because the dipole is aligned with the field:
τ = pE sin θ where θ = 0° => τ = 0
Final Conclusion: A non-uniform field exerts a net force on a dipole but no torque if aligned.

Quick Tip: In non-uniform fields, focus on field gradients for force and alignment for torque.

Solution: The electric field is zero outside the parallel plates (at P₁ and P₃) because the fields cancel. Inside the plates (at P₂), the electric field is uniform, and q experiences a force F₂ ≠ 0.
Final Conclusion: Outside the plates, the field cancels. Inside, the field is uniform, causing a net force.

Quick Tip: Always check the symmetry of the system to determine where fields cancel or add up.

Solution: When the spheres are brought into contact, charge redistributes in proportion to their capacitances. Capacitance of a sphere is proportional to its radius:
^Q₁⁄_Q2 = ^R₁⁄_R2
Final Conclusion: The charge ratio depends on the radii of the spheres directly.

Quick Tip: In conducting spheres, charge distribution is proportional to the size (radius) of the spheres.

Solution: Using Coulomb’s Law: F = k^q₁q₂⁄_r²
When both charges are doubled, q'₁ = 2q₁ and q'₂ = 2q₂, the force becomes:
F' = k ^{(2q₁)(2q₂)}⁄_r'²
F'=k^4q₁q₂⁄_r'² = 4F
To maintain the same force, the distance must increase by a factor of 2, as F ∝ 1/r². Hence, r' = 2d.
Final Conclusion: Doubling both charges requires doubling the distance to keep the force constant.

Quick Tip: Remember, the force is inversely proportional to the square of the distance. Any change in charge affects the force, and distance compensates accordingly.

Solution:
Step 1: Series Connection Properties
In a series combination:
1. The same charge Q flows through both capacitors.
2. The potential difference divides inversely proportional to capacitance:
V₁ = ^Q⁄_C1, V₂ = ^Q⁄_C2 => ^V₁⁄_V2 = ^C₂⁄_C1
Substitute C₁ = 2µF and C₂ = 3µF:
^V₁⁄_V2 = ^C₂⁄_C1 = ³⁄₂

Step 2: Energy Stored in Capacitors:
The energy stored in a capacitor is:
U= ¹⁄₂ CV²
For capacitors in series, since Q is the same:
U₁ = ^Q2⁄_2C1, U₂ = ^Q2⁄_2C2
The ratio of energies is:
^U₁⁄_U2 = ^C₂⁄_C1 = ³⁄₂
Conclusion: Both the potential ratio ^V₁⁄_V2 and the energy ratio ^U₁⁄_U2 are ³⁄₂.

Quick Tip: For capacitors in series, voltages divide inversely proportional to capacitance, and energy follows the same ratio.

Solution:
1. Total Resistance in Series: The total resistance of the series circuit is:
R_total = 100 + 200 = 300 Ω.

2. Current Through the Circuit: Using Ohm's Law, the current in the circuit is:
I = ^{Total Voltage}⁄_{Total Resistance} = ²⁰⁄₃₀₀ = ¹⁄₁₅ A.

3. Voltage Across the 200Ω Resistor: The voltage across the 200Ω resistor is:
V = I * R = ¹⁄₁₅ * 200 = ²⁰⁰⁄₁₅ = ⁴⁰⁄₃V This is approximately 13.33V, But in ideal case, Voltmeter will not draw any current. So, 10V is the Potential Difference.

4. Voltmeter Reading: Since the voltmeter is connected across the 200Ω resistor, its reading is equal to the voltage drop across this resistor, which is 10 V(In ideal case, where Voltmeter resistance is infinity).
Conclusion: The voltmeter reading across the 200Ω resistor is 10 V.

Quick Tip: In series circuits, the voltage across a resistor is proportional to its resistance. Use Ohm's Law to calculate voltage drops.

Solution: The equivalent emf of the parallel combination of the cells is:
E_eq = ^{E₁r₂ + E₂r₁}⁄_{r1 + r2} = ^{2 * 2 + 1 * 1}⁄_{1 + 2} = ⁵⁄₃ V
The equivalent resistance of the parallel cells is:
r_eq = ^r₁r₂⁄_{r1 + r2} = ^{1 * 2}⁄₁₊₂ = ²⁄₃Ω
Total resistance is R = r_eq + ⁴⁄₃ = 2Ω. Current is:
I= ^E_eq⁄_R = ^5/3⁄₂= ⁵⁄₆A
Final Conclusion: Use equivalent emf and resistance for combined cells to calculate current.

Quick Tip: For cells in parallel, calculate equivalent emf and internal resistance first.

Solution: Stretching the wire reduces its radius to half, increasing its length to four times. Resistance is:
R' = ρ^L'⁄_A' = ρ ^4L⁄_π(r/2)²
R' = 16R
Resistivity does not change, and power rating at constant voltage becomes:
P' = ^V²⁄_R' = ^P⁄₁₆
Final Conclusion: Stretching affects geometry and resistance but not resistivity.

Quick Tip: Changes in wire dimensions impact resistance proportionally; resistivity remains material-specific.

Solution: Diamagnetic materials have the lowest susceptibility, followed by paramagnetic, and ferromagnetic with the highest.
Final Conclusion: Susceptibility increases from diamagnetic to ferromagnetic.

Quick Tip: Magnetic susceptibility depends on the alignment of atomic dipoles in the material.

Solution: The force per unit length between two parallel current-carrying conductors is given by:
F/L = ^{μ₀I₁I₂}⁄_2πd
For a length L, the force is:
F= ^{μ₀I₁I₂L}⁄_2πd
Thus, the force is directly proportional to I₁, I₂ and L.
Final Conclusion: The force between parallel conductors depends on their current product and length.

Quick Tip: Memorize the formula for force per unit length for parallel currents. Always check dependence on distance d.

Solution:
1. Current Distribution: The current splits inversely proportional to the resistances of the two paths:
I_ABC = ^R⁄_3R * 3I = I, I_ADC = ^2R⁄_3R * 3I = 2I.

2. Magnetic Field Contribution: For a semicircular current-carrying wire, the magnetic field at the center is:
B= ^μ₀I⁄_4r
-The semicircle ABC contributes a field B_ABC = ^μ₀I_ABC⁄_4r = ^μ₀I⁄_4r.
- The semicircle ADC contributes a field B_ADC= ^μ₀I_ADC⁄_4r = ^μ₀(2I)⁄_4r = ^μ₀I⁄_2r .

3. Net Magnetic Field: The magnetic fields due to both semicircles add vectorially. Since the current flows in opposite directions in the two semicircles, their contributions will oppose. Hence,
B_net = |B_ABC - B_ADC| = | ^μ₀I⁄_4r - ^μ₀2I⁄_4r|
B_net = ^μ₀I⁄_4r
4. Direction: Using the right-hand rule for both semicircles, the net magnetic field is out of the plane, since B_ADC is greater.
Conclusion: The net magnetic field = ^μ₀I⁄_4r out of the plane.

Quick Tip: For semicircular loops, calculate the magnetic field contribution for each section and apply the right-hand rule for direction.

Solution: The torque experienced by a current loop in a magnetic field is given by:
τ = NIAB sin θ,
where:

• N = 1 (number of turns)
• I = 10 A (current in the loop)
• A = l² = (0.01)² = 10^-4 m² (area of the square loop)
• B = 0.2 T (magnetic field)
• θ = 90° (angle between the field and the normal to the plane of the loop, since the field is parallel to the plane, the normal to the loop makes 90° with the field).

Quick Tip: Torque on a current loop is maximized when the magnetic field is perpendicular to the plane and zero when parallel to the loop's normal.

Solution: In a purely capacitive circuit, current leads the voltage by 90° or π/2. This is a characteristic phase difference.
Final Conclusion: Phase difference of π/2 indicates purely capacitive behavior.

Quick Tip: For inductors, current lags voltage; for capacitors, current leads voltage.

Solution: Mutual inductance is given by:
M = ^ΔΦ⁄_ΔI
Here, ΔΦ = 15 Wb, ΔI = 10 A, so:
M= ¹⁵⁄₁₀ = 1.5H
Final Conclusion: Mutual inductance relates flux change to current change.

Quick Tip: Mutual inductance depends on flux linkage and current change. Always check units.

Solution:
Figure (a):

• The wire transforms into a circular loop, increasing the area.
• The magnetic field is into the plane.
• As the area increases, the magnetic flux through the loop also increases.
• To oppose this increase, the induced current flows in an anticlockwise direction, generating a magnetic field out of the plane.

• The wire transforms into a narrow straight shape, reducing the area.
• The magnetic field is out of the plane.
• As the area decreases, the magnetic flux through the loop decreases.
• To oppose this decrease, the induced current flows in an anticlockwise direction, generating a magnetic field out of the plane.

Quick Tip: Apply Lenz's Law systematically to analyze changes in flux and the direction of induced current.