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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 500 turns per meter carries a current of \( 4.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**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. Magnetic intensity H = n I . Given: n = 500 m⁻¹ , I = 4.5 A . Substitute: H = 500 × 4.5 = 2250 A m⁻¹ . Substituting values gives 2250 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 400 turns per meter carries a current of \( 5 \, \text{A} \). What is the magnetic intensity \( H \) ins

**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. Magnetic intensity H = n I . Given: n = 400 m⁻¹ , I = 5 A . Substitute: H = 400 × 5 = 2000 A m⁻¹ . Substituting values gives 2000 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 700 turns per meter carries a current of \( 3 \, \text{A} \). What is the magnetic intensity \( H \) ins

**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. Magnetic intensity H = n I . Given: n = 700 m⁻¹ , I = 3 A . Substitute: H = 700 × 3 = 2100 A m⁻¹ . Substituting values gives 2100 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 400 turns per meter carries a current of \( 6 \, \text{A} \). What is the magnetic intensity \( H \) ins

**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. Magnetic intensity H = n I . Given: n = 400 m⁻¹ , I = 6 A . Substitute: H = 400 × 6 = 2400 A m⁻¹ . Substituting values gives 2400 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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PHYSICS

A copper wire carries 4.5 A with a drift speed of 1.8 × 10⁻⁴ m/s . If n = 8.5 × 10²⁸ m^{-3 and e = 1.6 × 10⁻

Given: A copper wire carries 4.5 A with a drift speed of 1.8 × 10⁻⁴ m/s . If n = 8.5 × 10²⁸ m^{-3 and e = 1.6 × 10⁻¹⁹ C, what is the cross-sectional area? These values define the system as per NCERT data. Formula: Drift speed: v_d = I/n e A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Rearrange: A = I/n e v_d . Substitute: A = frac4.58.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.8 × 10⁻⁴. Calculate: A = 4.5/2.448 × 10⁵ approx 1.84 × 10⁻⁵ m² . 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A wire of length 1.8 m carrying 2 A is at 30° to a magnetic field of 0.45 T . What is the force on the wire?

Given: A wire of length 1.8 m carrying 2 A is at 30° to a magnetic field of 0.45 T . What is the force on the wire? These values define the system as per NCERT data. Formula: Force F = I l B sin θ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: F = 2 × 1.8 × 0.45 × sin 30° = 3.6 × 0.45 × 0.5 = 0.81 N . 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

What is the key factor determining the maximum current a battery can supply?

Maximum current occurs when external resistance is zero, so I_{max = ε / r . The internal resistance ( r ) limits the current, as it’s the only resistance in the circuit under short-circuit conditions.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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Acceleration and Velocity Analysis

A boat travels west at 6m/s while a current flows south at 8m/s. What is the magnitude of the boat’s velocity relative t

Velocity components: vx=−6m/s,vy=−8m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 10 m/s as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8

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