#square loop

Energy Stored in Inductor and Magnetic Energy

A square loop of side 20 cm rotates at 20 rad/s in a 0.15 T field. What is the maximum emf induced?

**Energy in inductor** cannot change instantaneously because that would require infinite power, current through inductor continuous, voltage may jump, principle used in chokes, inductive kick, back emf, explaining why inductor opposes change in current. A = (0.2)² = 0.04 m² . ε₀ = N B A ω = 1 × 0.15 × 0.04 × 20 = 0.12 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 = ½ L I², result 0.12 V follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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Rotational EMF and AC Generator

A square loop of side 18 cm rotates at 10 rad/s in a 0.25 T field. What is the maximum emf induced?

**Rotational emf** when coil area A rotates with angular speed ω in uniform field B, flux Φ = B A cos ωt, emf e = -N dΦ/dt = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = ω/2π. For square side 22 cm area 0.0484 m² N=1 ω=14 rad/s B=0.15 T, e₀=1×0.15×0.0484×14=0.1016 V, sinusoidal. A = (0.18)² = 0.0324 m² . ε₀ = N B A ω = 1 × 0.25 × 0.0324 × 10 = 0.081 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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Rotational EMF and AC Generator

A square loop of side 20 cm is rotated in a 0.2 T field at 10 rad/s. What is the maximum emf induced?

**Rotational emf** when coil area A rotates with angular speed ω in uniform field B, flux Φ = B A cos ωt, emf e = -N dΦ/dt = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = ω/2π. For square side 22 cm area 0.0484 m² N=1 ω=14 rad/s B=0.15 T, e₀=1×0.15×0.0484×14=0.1016 V, sinusoidal. ε₀ = N B A ω , N = 1 , A = (0.2)² = 0.04 m² . ε₀ = 1 × 0.2 × 0.04 × 10 = 0.08 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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Rotational EMF and AC Generator

A square loop of side 15 cm rotates at 15 rad/s in a 0.2 T field. What is the maximum emf induced?

**AC generator** principle same as rotating coil, N=200 turns A=0.04 m² B=0.1 T f=50 Hz ω=2π×50=314 rad/s, e₀= N B A ω =200×0.1×0.04×314=251.2 V, emf e= e₀ sin ωt, frequency equals rotation frequency, maximum when plane parallel to field. A = (0.15)² = 0.0225 m² . ε₀ = N B A ω = 1 × 0.2 × 0.0225 × 15 = 0.0675 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 = ½ L I², result 0.0675 V follows,

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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Torque on Current Loop, Magnetic Moment and Galvanometer

A square loop of side \( 0.15 \, \text{m} \) with 50 turns carries \( 1 \, \text{A} \) in a magnetic field of \( 0.6 \,

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Torque tau = N I A B sin θ , where A = 0.15 × 0.15 = 0.0225 m² . tau = 50 × 1 × 0.0225 × 0.6 × sin 30° = 0.675 × 0.5 =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.25 \, \text{m} \) with 20 turns carries \( 2.5 \, \text{A} \) in a magnetic field of \( 0.3 \

**Field at centre of circular loop** with N turns is B = μ₀ N I/(2R), R radius (m), direction along axis via right-hand rule, magnitude proportional to N I/R. For R = 0.09 m, N = 45, I = 1.2 A, B = 4π×10⁻⁷×45×1.2/(2×0.09) = 3.77×10⁻⁴ T, showing N enhancement. Torque tau = N I A B sin θ , where A = 0.25 × 0.25 = 0.0625 m² . tau = 20 × 2.5 × 0.0625 × 0.3 × sin 45° = 0.9375 × 0.707 = 0.6633 ≈ 0.66 N m . Using F = q v B sinθ, F = I l B