#perpendicular motion

Self-Induction and Self-Inductance

A conducting rod moves perpendicular to a uniform magnetic field with constant velocity. What is true about the induced

**Self-induction** emf induced in coil due to change in its own current, e = -L dI/dt, L self-inductance (H), L = μ₀ N² A / l for solenoid, N turns, A area (m²), l length (m), μ₀=4π×10⁻⁷ H/m. For solenoid 650 turns/m means n=650, A=0.014 m², L = μ₀ n² A l? Actually per unit length? For length l, N=n l, L= μ₀ n² A l, if l=1 m, L=4π×10⁻⁷×650²×0.014=7.43×10⁻³ H, dI/dt=(3-6)/0.25=-12 A/s, e= -L×(-12)=0.089 V. The induced emf ( ε = B l v ) remains constant because the velocity, magnetic field, and rod length are constant, leading to a steady flux change rate.

Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance

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Motional EMF - Rod and Rectangular Loop

A rod of length 0.4 m moves at 2 m/s in a 0.3 T field perpendicular to its length. What is the induced emf?

**Loop sides 35 cm and 15 cm** moving out B=0.8 T v=1.5 m/s perpendicular to shorter side 15 cm, so cutting side =35 cm=0.35 m? Actually motion perpendicular to shorter side means longer side cuts, e= B×(long side)×v =0.8×0.35×1.5=0.42 V, illustrating motional emf e = B L v. ε = B l v = 0.3 × 0.4 × 2 = 0.24 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 0.24 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 7 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.2 \, \text{T} \). What is the ma

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 7 × 10⁶ × 0.2 = 2.24 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 2.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.5 \, \text{T} \). What is the

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. r = (mv/qB) . r = (9.1 × 10⁻³¹ × 2.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (2.275 × 10⁻²⁴/8 × 10⁻²⁰) = 2.84375 × 10⁻⁵ m ≈ 2.84 × 10⁻³ cm . Using F = q v B sinθ, F = I l B sinθ, B

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

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