#magnetic field calculation

Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 1.2 \, \text{A m}^2 \) produces a field at \( 0.3 \, \text{m} \) on its equatorial line. What i

**Field due to bar magnet** on axial line is B_axial = (μ₀/4π)·2m/r³, equatorial B_eq = (μ₀/4π)·m/r³, where μ₀/4π = 10⁻⁷ T·m/A, m magnetic moment (A·m²), r distance (m). Axial field twice equatorial at same distance and parallel to moment, equatorial opposite to moment direction. B = (μ₀/4π) (m/r³) . Given: m = 1.2 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (1.2/(0.3)³) = 10⁻⁷ × (1.2/0.027) ≈ 4.44 × 10⁻⁶ T . Substituting values gives 4.44 × 10⁻⁶ T, 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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment \( 2 \, \text{A m}^2 \

**Field due to bar magnet** on axial line is B_axial = (μ₀/4π)·2m/r³, equatorial B_eq = (μ₀/4π)·m/r³, where μ₀/4π = 10⁻⁷ T·m/A, m magnetic moment (A·m²), r distance (m). Axial field twice equatorial at same distance and parallel to moment, equatorial opposite to moment direction. The magnetic field on the equatorial line is B = (μ₀/4π) (m/r³) . Given: m = 2 A m² , r = 0.1 m , (μ₀/4π) = 10⁻⁷ T m A⁻¹ . Substitute: B = 10⁻⁷ × (2/(0.1)³) = 10⁻⁷ × (2/0.001) = 2 × 10⁻⁴ T . Substituting values gives 2 × 10⁻⁴ T, which matches expected magnitude for this

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

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Torque on Magnetic Dipole and Potential Energy

The magnetic field contribution \( B_m \) due to a material with \( M = 3 \times 10^5 \, \text{A m}^{-1} \) is: (Take \(

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. B_m = μ₀ M . Given: M = 3 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 3 × 10⁵ = 0.3768 T ≈ 0.38 T . Substituting values gives 0.38 T, 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 > Torque on Magnetic Dipole and Potential Energy

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PHYSICS

A bar magnet of magnetic moment 0.5 A m² is placed at a distance of 20 cm from its nter along its axis. Calculate the m

Given: A bar magnet of magnetic moment 0.5 A m² is placed at a distance of 20 cm from its nter along its axis. Calculate the magnetic field B at that point. (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: The magnetic field along the axis of a bar magnet is given by B = μ_0/4π 2m/r³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.5 A m², r = 20 cm = 0.2 m, μ_0/4π = 10⁻⁷ T m A^{-1 . Substitute: B = 10⁻⁷ × 2 × 0.5/(0.2)³ = 10⁻⁷ × 1/0.008 = 10⁻⁷ × 125 = 1.25 × 10⁻⁵ T . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A bar magnet with m = 1.5 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4Ï€ × 10⁻â

Given: A bar magnet with m = 1.5 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: B = μ_0/4π m/r³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 1.5 A m², r = 0.5 m, μ_0/4π = 10⁻⁷. B = 10⁻⁷ × 1.5/(0.5)³ = 10⁻⁷ × 1.5/0.125 = 1.2 × 10⁻⁶ T . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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