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#field calculation

3 public questions tagged with this topic.

Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet produces a field of \( 4 \times 10^{-6} \, \text{T} \) at \( 0.2 \, \text{m} \) on its equatorial line. Wha

**Axial versus equatorial** field comparison shows B_axial = 2 B_eq for same r. Using μ₀ = 4π×10⁻⁷ T·m/A, calculation involves r³ = (0.3)³ = 0.027 m³, so B = 10⁻⁷·m/r³ yields moment estimation. B = (μ₀/4π) (m/r³) , so m = (B r³/(μ₀/4π)) . Given: B = 4 × 10⁻⁶ T , r = 0.2 m , (μ₀/4π) = 10⁻⁷ . m = (4 × 10⁻⁶ × (0.2)³/10⁻⁷) = (4 × 10⁻⁶ × 0.008/10⁻⁷) = 0.32 A m² . Substituting values gives 0.32 A m², which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Continuous Charge Distribution

An infinite line charge produces an electric field of \( 1.8 \times 10^5 \, \text{N/C} \) at 3 cm. What is its linear ch

**Line charge concept** extends point charge to infinite wire where symmetry dictates radial field proportional to λ and inversely proportional to distance r. λ = q/L for uniform case, field direction depends on sign of λ, outward for positive. E = (λ/2 π ε₀ r) , or E = (2 k λ/r) , k = 9 × 10⁹ N·m²/C² . 1.8 × 10⁵ = (2 × 9 × 10⁹ × λ/0.03) . λ = (1.8 × 10⁵ × 0.03/18 × 10⁹) = 3 × 10⁻⁷ C/m . Substituting values gives 3.0 × 10⁻⁷ C/m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and

Ref: NCERT > Physics Book > Electric Charges and Fields > Continuous Charge Distribution

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Coulomb's Law and Force Between Point Charges

A point charge \( -10 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 10 m along the

**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 10 × 10⁻⁶ C , r = 10 m . E = 9 × 10⁹ × (10 × 10⁻⁶/(10)²) = 9 × 10⁹ × (10 × 10⁻⁶/100) = 9 × 10² N/C . Substituting values gives 900 N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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