#turns per metre

Self-Induction and Self-Inductance

A solenoid of 850 turns/m and area 0.01 m² has \( \mu_r = 1 \). What is its self-inductance? (\( \mu_0 = 4\pi \times 10^

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. L = μ_r μ₀ n² A l , assume l = 1 m . L = 1 × 4π × 10⁻⁷ × (850)² × 0.01 × 1 = 0.00907 H ≈ 0.009 H . 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

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

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Solenoid, Toroid and Ampere's Law

A solenoid has 850 turns per meter and carries a current of \( 1.6 \, \text{A} \). What is the magnetic field inside it?

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 850 × 1.6 = 5.44 π × 10⁻⁴ ≈ 1.71 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A solenoid with 1800 turns per meter carries a current of \( 1.5 \, \text{A} \). What is the magnetic field inside it? (

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 1800 × 1.5 = 10.8 π × 10⁻⁴ ≈ 3.39 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A solenoid with 900 turns per meter carries \( 2 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi \t

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. B = μ₀ n I . B = 4 π × 10⁻⁷ × 900 × 2 = 7.2 π × 10⁻⁴ ≈ 2.26 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A solenoid with 2500 turns per meter carries a current of \( 2 \, \text{A} \). What is the magnetic field inside it? (\(

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 2500 × 2 = 20 π × 10⁻⁴ ≈ 6.28 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A solenoid with 2000 turns per meter carries \( 0.8 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. B = μ₀ n I . B = 4 π × 10⁻⁷ × 2000 × 0.8 = 6.4 π × 10⁻⁴ ≈ 2.01 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 900 turns per meter and current \( 3.5 \, \text{A} \) has a core with \( \mu_r = 150 \). What is \( B \)

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. B = μ₀ μ_r n I . Given: n = 900 m⁻¹ , I = 3.5 A , μ_r = 150 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 150 × 900 × 3.5 = 0.59346 T ≈ 0.59 T . Substituting values gives 0.59 T, which matches expected magnitude

Ref: NCERT > Physics Book > Magnetism and Matter > Solenoid with Magnetic Core and Magnetic Properties

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