Convert speed to m/s: 1080×(5/18) = 300 m/s

EMF = Bvl = (1.75×10⁻⁵)(300)(40)

is equal to 0.21V

Induced EMF e = -M(di/dt)

di/dt = i_m ω cos ωt

Maximum EMF = Mi_m ω = 0.005×10×100π = 5π

When isolated, charge Q remains constant as there's no path for charge flow

Capacitance C = εₒA/d decreases when d increases, as C ∝ 1/d

Since V = Q/C and Q is constant, V must increase when C decreases

Inside shell potential constant

Equal to surface potential

Outside decreases as 1/r

Field is radially outward

Dipole moment is axial

Angle between them is 180°

Initial force F ∝ q₁q₂

New charges: 3μC and 6μC

New force = (3×6)/(6×9)×F = F/3

Electric field lines start from +ve charge

End at -ve charge or infinity

Cannot form closed loops

Use conservation of energy: initial PE = final KE

Calculate PE change: ΔU = k(q₁q₂)(1/r₂ - 1/r₁)

Find velocity: (1/2)mv² = kq₁q₂(1-0.1) → v = 180 ms⁻¹

Consider area swept by each spoke

Calculate average velocity: vavg = ωL/2

Induced emf = BLvavg = (1/2)ωBL²

Consider electric field lines: radially outward from point charge

Equipotential surfaces are perpendicular to field lines

Must be concentric spheres centered at charge

Write expressions: Eax = (2p/4πε₀r³) and Eeq = (p/4πε₀r³)

Compare magnitudes: Eax = 2Eeq

Consider directions: Eax is opposite to Eeq

Apply Gauss's law: ∮E.ds = q/ε₀ for closed surface

For outside charge, flux entering = flux leaving, net flux = 0

For inside charge, net flux = q/ε₀

Apply F = eE = ma for electric force

Use equation of motion: h = (1/2)at², where a = eE/m

Solve for t: t = √(2hm/eE)

Use motional EMF formula: ε = Blv sinθ

Consider vertical component: B cosθ

Calculate: ε = 4×10⁻⁴×20×400×sin30° = 1.6 V

Consider brake mechanism

Eddy currents induced in tracks

These currents produce opposing force

For minimum capacitance, connect capacitors in series

Apply series formula: 1/C = 1/1 + 1/2 + 1/4

Solve to get C = 4/7 pF

Calculate equivalent capacitance: 1/C = 1/2 + 1/4, C = 4/3μF

Find charge: Q = CV = (4/3)×6 = 8μC

Energy = ½QV = ½×8×6 = 24μJ

Calculate initial potential energy: U₁ = kq₁q₂/r₁

Calculate final potential energy: U₂ = kq₁q₂/r₂

Work done = U₂ - U₁ = 1.35×10⁻⁷ J

Calculate total charge: Q = 4×10¹⁰ × 1.6×10⁻¹⁹ C

Use electric field formula: E = kQ/r²

Substitute values with k = 9×10⁹ Nm²/C²

Use Faraday's law: ε = -dΦ/dt

Differentiate Φ: dΦ/dt = 6t + 4

Substitute t = 2: ε = 16V

Initial force: F = kq₁q₂/d² = k(2×-8)/d² = -16k/d²

After contact: Each sphere has -3nC (total charge divided equally)

For same force magnitude: k(2×8)/d² = k(3×3)/(x²); solve for x = 3d/4

Convert speed to m/s: 3600 km/h = 1000 m/s

Use e.m.f. formula: ε = Bℓv sin θ, where B_vertical = B sin(30°)

Calculate: ε = 4×10⁻⁴ × 25 × 1000 × sin(30°) = 5V

Calculate initial charge stored: For series combination, Q is same across both: Q = C₁V₁ = C₂V₂ and V = V₁ + V₂. Find Q using Q = V/(1/C₁ + 1/C₂)

When reconnected in parallel, this charge redistributes. Total charge remains constant. For parallel combination, V is same across both capacitors

Calculate final voltage: V = Q/(C₁ + C₂) = 200V

Recall definition of equipotential surface: all points at same potential

Note that work done = qΔV

Since ΔV = 0 on equipotential surface, work done = 0

Use work-energy principle: qV = ½mv²

Substitute q = 2C, V = 1V, m = 1kg

Solve for v: v = √(4/1) = 2 ms⁻¹

Calculate effective distance considering dielectric: r_eff = r/2 + (r/2)/√K

Use Coulomb's law with modified distance

Find ratio of new force to original force

Use Coulomb's law for electric field: E = kq/r² where k = 9×10⁹ Nm²/C²

Put E = 2 N/C, r = 0.3m

Solve for q: q = Er²/k = 2(0.3)²/9×10⁹ = 2×10⁻¹¹ C

Write dimensions of flux (φ = BA)

Write dimensions of permeability (μ = B/H)

Divide flux dimensions by permeability

Total potential difference = 30V - (-10V) = 40V

Field = Potential difference/distance

E = 40V/0.02m = 2000 V/m

Total capacitance in parallel = 3C

Charge distributes according to capacitance ratio

Q1 = Q/3, Q2 = 2Q/3

For like charges, PE = kq₁q₂/r increases with r

For unlike charges, PE = -kq₁q₂/r decreases with r

Therefore behavior depends on charge signs

Non-uniform field creates net force due to different fields at + and - charges

Torque τ = p×E exists if dipole not aligned with field

Both force and torque present in non-uniform field

Electric force F = qE is constant

Work done = Force × distance = qEy

All work done converts to K.E. = qEy

Force F ∝ q₁q₂ where q₁ + q₂ = 2Q

For maximum force, q₁ = q₂ = Q

Therefore Q/q₁ = 1

Magnetic field in toroid B = μ₀nI where n = 500 turns/m

Energy density = B²/2μ₀ = μ₀n²I²/2

Substitute values: μ₀(500)²(2)²/2 = 0.628 J/m³

The electric field between the dees provides acceleration whenever particle crosses the gap. This happens twice per revolution

Inside dees, magnetic field only provides centripetal force without changing speed. No work is done by magnetic force

The particle gains energy continuously as it crosses the gap multiple times, making larger and larger orbits

Use Faraday's law for self-induction: induced emf = -L(di/dt). First find rate of change of current: Δi/Δt = (2-5)/0.5 = -6 A/s.

Multiply by inductance and take magnitude: emf = L di/dt = 0.2×6

Calculate: emf = 1.2 V. The negative sign in formula indicates emf opposes change, but question asks for magnitude.

When a bubble gets charged, all like charges spread across its surface. These charges experience mutual repulsion according to Coulomb's law.

The repulsive force between these like charges creates an additional outward force that works against the surface tension of the bubble.

This additional outward force causes the bubble to expand until it reaches a new equilibrium where the outward electrostatic force balances with the inward surface tension force.

For parallel combination: Cp = C + C = 2C. For series: Cs = CC/(C+C) = C/2. We're given Cp - Cs = 6μF

Substitute: 2C - C/2 = 6. Multiply all terms by 2: 4C - C = 12

Solve: 3C = 12, therefore C = 4μF

For small oscillations of dipole in electric field, restoring torque τ = -PE sinθ ≈ -PEθ for small θ. This is similar to SHM.

Compare with standard SHM equation: τ = -Iα where α is angular acceleration. This gives equation of motion: Iα = -PEθ

For SHM, period T = 2π√(I/PE), similar to T = 2π√(m/k) for linear SHM

Let's start with the formula for electric field due to infinite wire: E = λ/2πε₀r. First convert units - charge density to C/m and distance to meters.

Now substitute values: λ = (1/4×10⁻²) C/m, r = 0.2 m, ε₀ = 8.85×10⁻¹² C²/N⋅m²

Calculate: E = (0.25×10⁻²)/(2π×8.85×10⁻¹²×0.2) = 2.25×10⁸ N/C

For voltage requirement: need 3 capacitors in series (1500/500 = 3)

Each series group gives 2/3 µF. Need 9 parallel groups for 6µF

Total capacitors = 3×9 = 27

Polar molecules have permanent separation of charge centers even without field

External field can enhance separation and cause alignment

They DO possess permanent dipole moments - statement C is false

Capacitance of sphere C = 4πε₀r. Volume remains constant: V₁ = V₂

r_new = r_old×∛8 (volume relation)

Therefore, C_new = 2C_old (linear with radius)

Original capacity C₀ = ε₀A/d. With dielectric: C = Kε₀A/(d+x(1-1/K))

For same capacity: d = d+x(1-1/K)

Solve for K using given values: K = 8

Use work-energy theorem: ΔKE = qEx (only x-component matters)

0.12 = Q(300)(0.5) (convert units carefully)

Solving gives Q = 800µC

Rod rotation creates magnetic field. Moving charges in rod experience magnetic force

Force on charges qv×B = qωrB creates electric field E = F/q

Solving for E yields mω²x²/e (balancing centripetal force)

Use Gauss's law: E·2πrl = λ/ε₀, where l is length, λ is linear charge density

Solving for E: E = λ/(2πε₀r)

Therefore, E ∝ 1/r, matching option B

Let's first understand what wb A⁻¹ represents. This unit comes from dividing magnetic flux (weber) by current (ampere). In electromagnetic induction, whenever we divide flux by current, we're dealing with inductance.

Now, there are two types of inductance we need to consider. Self inductance (L) is defined as L = Φ/I, where Φ is the flux produced in the same coil. When we plug in the units, we get wb/A.

Similarly, for mutual inductance (M), we use M = Φ₂₁/I₁, where Φ₂₁ is the flux in second coil due to current I₁ in first coil. Again, this gives us wb/A. Therefore, both types of inductance are measured in wb A⁻¹.

Forces acting: weight component down slope (mgsinθ) and magnetic force (BIl) up slope

For equilibrium: BIl = mgsinθ. Mass/length = 0.5kg/m, so for unit length calculation is simpler

I = (0.5×10×sin30°)/(0.25) = 11.32A. Direction must oppose sliding!

Self inductance L = NΦ/I, where N=turns, Φ=flux per turn, I=current.

Substitute values: L = (500)(4×10⁻³)/2

This gives L = 1.0 henry. Units check: Wb/A = henry!

EMF is rate of change of magnetic flux. Flux = NBA where N=turns, B=field, A=area.

Initial flux = (80)(1.0)(0.01) = 0.8 Wb. Final flux = 0 Wb. Change happens in 0.2s.

EMF = -ΔΦ/Δt = -(0-0.8)/0.2 = 4V. Negative sign shows induced EMF opposes change!

When we insert a dielectric (glass), capacitance increases by factor K (dielectric constant). Since battery is disconnected, charge remains constant.

From Q=CV, if C increases and Q is constant, V must decrease.

Energy stored = ½CV² or Q²/2C. With C increased and Q constant, energy must decrease. Only potential difference and energy decrease!

At any point, total potential is sum of potentials due to individual charges. Distance to midpoint is 20cm for both charges.

For each charge: V = kQ/r. Here k=9×10⁹, Q₁=1.8×10⁻⁶C, Q₂=2.8×10⁻⁶C, r=0.2m

V = (9×10⁹)[(1.8×10⁻⁶)/0.2 + (2.8×10⁻⁶)/0.2] = 2.1×10⁵ V

Think about how we experience gravity without touching Earth, or how a compass needle aligns without touching a magnet. Fields explain these "action at a distance" forces!

Fields are not just mathematical tools - we can measure their effects. For example, iron filings show magnetic field patterns, and we can measure electric fields with test charges.

While fields do help explain some contact forces (like electromagnetic forces between atoms), their main purpose is explaining how objects can influence each other without touching.

Let's think about a dipole - it's two equal and opposite charges separated by a small distance. At a point far from the dipole, we need to consider both charges' effects. The field involves a vector sum while potential involves scalar sum.

For electric field, when we do the vector math, the leading terms cancel out (1/r²), leaving us with the next term, which goes as 1/r³. This makes sense as dipole field falls off faster than single charge field!

For potential, the leading terms in 1/r also cancel, leaving us with 1/r². Again, this falls off faster than single charge potential (1/r). This pattern - field falling off one power faster than potential - is consistent with E = -∇V.

In a uniform electric field, the force on a charged particle is constant (F = qE). This is similar to an object under constant acceleration! The force causes acceleration a = qE/m

Using equations of motion, velocity after time t is v = at = (qE/m)t. Now we're ready to find kinetic energy!

KE = ½mv² = ½m(qE/m)²t² = E²q²t²/2m. Notice how mass appears in denominator - lighter particles gain more KE for same field!

The torque on a dipole depends on both the dipole moment and electric field strength, but also on the angle between them. The formula is τ = pE sin θ. Let's identify our values: p = 4×10⁻⁹ C-m, E = 5×10⁴ N/C, θ = 30°

Now substitute these values: τ = (4×10⁻⁹)(5×10⁴)(sin 30°). Remember sin 30° = 0.5. Always keep track of units as we multiply!

This gives us τ = (4×10⁻⁹)(5×10⁴)(0.5) = 10⁻⁴ Nm. The units work out: (C-m)(N/C) = Nm, which is exactly what we want for torque!

Let's approach this systematically. At point O (center), we have forces from all four corners. Due to equal distance from center to corners, only magnitude of charges matters. Draw forces from each charge to center.

By symmetry, forces from charges +q on opposite corners cancel out their horizontal components, leaving only vertical components. The +2q and -2q charges create a stronger effect along BD diagonal.

The net force must be along BD because +2q attracts and -2q repels along this diagonal, while other forces' components cancel out. This is why the resultant is along BD diagonal.

Let's visualize what's happening: we have an oil drop falling under gravity but also experiencing an electric force and air resistance. At terminal velocity, all these forces balance. First, identify the forces: gravity (mg), electric force (qE), and air resistance (6πηrv).

At terminal velocity, the sum of forces is zero: mg - qE - 6πηrv = 0. We can find the radius using Stokes' law and terminal velocity. Using the given density of oil (900 kg/m³) and viscosity of air (1.8×10⁻⁵ Ns/m²), let's calculate the radius.

Once we have the radius, we can find mass and substitute everything into our force balance equation. The charge q comes out to be 8×10⁻¹⁹ C. Always check if this makes physical sense - it's a small charge, which is reasonable for an oil drop!

For straight wire: E₁=λ/2πε₀r

For semicircle: E₂=λ/2πε₀r

E₁=E₂

With battery connected, V remains constant

C₁=KC₀, C₂=C₀

Therefore C₁>C₂, V₁=V₂

For ideal transformer, no power loss

Power output = Power input

Verify no other factors affect power

Analyze induction process

Charges separate in sphere

Net negative after rod removal

Calculate velocity: v=gt=10×2=20m/s

Use e=Blv formula with given B, l

e=3×10⁻⁵×1×20=6×10⁻⁴V

Use Faraday's law: e = -L(di/dt)

Calculate di/dt = (5-2)/0.3 = 10A/s

L = e/(di/dt) = 1.0/10 = 0.1H = 100mH

Use Faraday's law: induced emf ε=-dΦ/dt

Apply Ohm's law: ε=IR=2×10=20V

Calculate flux change: ΔΦ=εΔt=20×10⁻³=2×10⁻² Wb

Recall Gauss's law relates flux to enclosed charge: Φ=Q_enclosed/ε₀

For dipole, total charge enclosed is zero as positive and negative charges are equal

Therefore, total flux must be zero regardless of radius

Apply Coulomb's law to find force on A due to B: F₁=k(10×20)×10⁻¹²/1² N

Apply Coulomb's law to find force on B due to A: F₂=k(20×10)×10⁻¹²/1² N

By Newton's third law, these forces are equal in magnitude but opposite in direction, so F₁=-F₂

Recall energy stored in capacitor U=½CV². For rate of change, use chain rule dU/dt

Apply chain rule: dU/dt=½C×2V×dV/dt

Calculate: dU/dt=5×10⁻⁶×4×0.6=12×10⁻⁶ W=12µW

Convert -3.2µC to Coulomb: -3.2×10⁻⁶ C. Negative charge means excess electrons present

Use relation Q=ne where e=1.6×10⁻¹⁹ C is elementary charge. Rearrange to get n=Q/e

Calculate n = 3.2×10⁻⁶/1.6×10⁻¹⁹ = 2×10¹³ electrons

Understand that in electrostatic equilibrium, charges must be stationary

In a conductor, free electrons move until reaching equilibrium where internal field is zero

All excess charge must reside on surface to maintain zero internal field

Convert dimensions to meters: 10cm×30cm = 0.1m×0.3m = 0.03m²

Apply electric flux formula Φ=E·A. Since plane is parallel to Z-X plane, it's perpendicular to field (Y-axis)

Calculate Φ=EA=(3×10⁵)(0.03)=9×10³Vm