#current calculation

Solenoid with Magnetic Core and Magnetic Properties

A solenoid with magnetic field \( B = 0.8 \, \text{T} \) inside has a core with \( \mu_r = 400 \) and \( n = 800 \, \tex

**Solenoid with magnetic core** produces field B = μ₀ μ_r n I inside, μ₀ = 4π×10⁻⁷ T·m/A, μ_r relative permeability, n = N/L turns per meter, I current. Core enhances field μ_r times, so given B, μ_r, n, current I = B/(μ₀ μ_r n) can be found, illustrating core effect on field strength. B = μ₀ μ_r n I , so I = (B/μ₀ μ_r n) . Given: B = 0.8 T , μ_r = 400 , n = 800 m⁻¹ , μ₀ = 4π × 10⁻⁷ . Substitute: I = (0.8/4π × 10⁻⁷ × 400 × 800) = (0.8/4π × 3.2 × 10⁻²) ≈

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 1.2 \, \text{T} \) with a core of \( \mu_r = 400 \) and \( n = 1500 \, \text{m}^{-1} \). What

**Core magnetization** M = (μ_r -1)nI, so B = μ₀(nI + M). High μ_r materials like soft iron increase B dramatically for same nI, used in electromagnets, with μ_r up to 5000, enabling strong fields with low current. B = μ₀ μ_r n I , so I = (B/μ₀ μ_r n) . Given: B = 1.2 T , μ_r = 400 , n = 1500 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (1.2/4π × 10⁻⁷ × 400 × 1500) = (1.2/7.539 × 10⁻¹) ≈ 1.592 A ≈ 1.6 A . Substituting values gives 1.6 A, which matches expected magnitude for this magnetic

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 0.9 \, \text{T} \) with a core of \( \mu_r = 300 \) and \( n = 1500 \, \text{m}^{-1} \). What

**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 , so I = (B/μ₀ μ_r n) . Given: B = 0.9 T , μ_r = 300 , n = 1500 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (0.9/4π × 10⁻⁷ × 300 × 1500) = (0.9/5.654 × 10⁻¹) ≈ 1.59 A ≈ 1.6 A .

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 0.96 \, \text{T} \) with a core of \( \mu_r = 300 \) and \( n = 1000 \, \text{m}^{-1} \). Wha

**Solenoid with magnetic core** produces field B = μ₀ μ_r n I inside, μ₀ = 4π×10⁻⁷ T·m/A, μ_r relative permeability, n = N/L turns per meter, I current. Core enhances field μ_r times, so given B, μ_r, n, current I = B/(μ₀ μ_r n) can be found, illustrating core effect on field strength. B = μ₀ μ_r n I , so I = (B/μ₀ μ_r n) . Given: B = 0.96 T , μ_r = 300 , n = 1000 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (0.96/4π × 10⁻⁷ × 300 × 1000) = (0.96/3.769 × 10⁻¹) ≈ 2.548 A ≈

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 0.6 \, \text{T} \) with a core of \( \mu_r = 200 \) and \( n = 1000 \, \text{m}^{-1} \). What

**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 , so I = (B/μ₀ μ_r n) . Given: B = 0.6 T , μ_r = 200 , n = 1000 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (0.6/4π × 10⁻⁷ × 200 × 1000) = (0.6/2.513 × 10⁻¹) ≈ 2.39 A ≈ 2.4 A .

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 1.2 \, \text{T} \) with a core of \( \mu_r = 400 \) and \( n = 2000 \, \text{m}^{-1} \). What

**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 , so I = (B/μ₀ μ_r n) . Given: B = 1.2 T , μ_r = 400 , n = 2000 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (1.2/4π × 10⁻⁷ × 400 × 2000) = (1.2/1.005 × 10⁰) ≈ 1.19 A ≈ 1.2 A .

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Variation of Conductivity with Concentration and Measurement

What is the time (in seconds) required to deposit 0.355 g of cobalt from a CoSO₄ solution using a current of 0.2 A? (Ato

Co²⁺ + 2e⁻ → Co . 1 mol Co (59 g) requires 2F. Moles = (0.355/59) = 0.006017 mol , Charge = 0.006017 × 2 × 96500 = 1161.24 C . t = (Q/I) = (1161.24/0.2) = 5806.2 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Electrolytic Cells and Electrolysis and Faraday's Laws

A dry cell delivers 0.25 A for 9650 s. How many grams of zinc are oxidized at the anode? (Atomic mass of Zn = 65 g/mol,

Charge = 0.25 × 9650 = 2412.5 C . Zn → Zn²⁺ + 2e⁻ , 1 mol Zn (65 g) requires 2F. Faradays = (2412.5/96500) = 0.025 F , Moles = (0.025/2) = 0.0125 mol , Mass = 0.0125 × 65 = 0.8125 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the time (in seconds) required to deposit 0.585 g of chromium from a Cr₂(SO₄)₃ solution using a current of 0.5 A

Cr³⁺ + 3e⁻ → Cr . 1 mol Cr (52 g) requires 3F. Moles = (0.585/52) = 0.01125 mol , Charge = 0.01125 × 3 × 96500 = 3256.875 C . t = (Q/I) = (3256.875/0.5) = 6513.75 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the time (in seconds) required to deposit 0.585 g of chromium from a Cr₂(SO₄)₃ solution using a current of 0.5 A

Cr³⁺ + 3e⁻ → Cr . 1 mol Cr (52 g) requires 3F. Moles = (0.585/52) = 0.01125 mol , Charge = 0.01125 × 3 × 96500 = 3256.875 C . t = (Q/I) = (3256.875/0.5) = 6513.75 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A current of 1 A deposits 0.635 g of Cu from CuSO₄ in 1930 s. What is the time required to deposit 0.54 g of Al from Al₂

Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A current of 1 A deposits 0.635 g of Cu from CuSO₄ in 1930 s. What is the time required to deposit 0.54 g of Al from Al₂

Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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