#JEE physics

Motional EMF - Rod and Rectangular Loop

A rectangular loop of 0.18 m × 0.3 m moves out of a 0.25 T field at 1 m/s along its shorter side. What is the emf?

**Motional emf** for rod length l moving with velocity v perpendicular to uniform field B, e = B l v (V), B in T, l in m, v in m/s, direction given by right-hand rule, positive end where positive charges accumulate due to q v×B force. For l=0.9 m, v=1.2 m/s, B=0.5 T, e=0.5×0.9×1.2=0.54 V. ε = B l v , l = 0.3 m . ε = 0.25 × 0.3 × 1 = 0.075 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l,

Ref: NCERT > Physics Book > Electromagnetic Induction > Motional EMF - Rod and Rectangular Loop

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Induced EMF Due to Change in Magnetic Field

A coil of 140 turns and area 0.05 m² is in a 0.1 T field that drops to zero in 0.25 s. What is the induced emf?

**Uniform field change** in coil produces emf proportional to area and turns, for circular coil radius 0.16 m area πr²=0.0804 m², B 0.12 T deformed to wire in 0.6 s, ΔΦ=0.12×0.0804=0.00965 Wb, e=0.00965/0.6=0.0161 V, illustrating area change also induces emf. Δ Φ = B A = 0.1 × 0.05 = 0.005 Wb . ε = N (Δ Φ/Δ t) = 140 × (0.005/0.25) = 140 × 0.02 = 2.8 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U =

Ref: NCERT > Physics Book > Electromagnetic Induction > Induced EMF Due to Change in Magnetic Field

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Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 5 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. Work done: W = p E (cos θ₀ - cos θ₁) = 5 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 180°) . W = 5 × 10⁻⁹ × 2 × 10⁵ × (0 - (-1)) = 5 × 10⁻⁹ × 2 × 10⁵ × 1 = 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Potential Energy of System of Charges

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 2 \t

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. U = -p E cos θ = -6 × 10⁻⁹ × 2 × 10⁵ × cos 45° . cos 45° = (1/√(2)) ≈ 0.707 , so U = -6 × 10⁻⁹ × 2 × 10⁵ × 0.707 = -8.48 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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