#magnetism

CHEMISTRY

What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ?

Given: What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ? These values define the system as per NCERT data. Formula: Magnetic moment = sqrt2(2+2) = sqrt8 approx 2.83 BM.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Ni²⁺ ( d⁸ ) with NH₃ (moderate field) in an octahedral field is high spin ( t_{2g⁶ e_g² ), with 2 unpaired electrons. 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 Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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PHYSICS

A material with susceptibility chi = -3 × 10⁻⁵ has a relative permeability μ_r of:

Given: A material with susceptibility chi = -3 × 10⁻⁵ has a relative permeability μ_r of: These values define the system as per NCERT data. Formula: μ_r = 1 + chi. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: chi = -3 × 10⁻⁵. Substitute: μ_r = 1 - 3 × 10⁻⁵= 0.99997 . 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 material with susceptibility chi = 3 × 10⁻⁴ has a relative permeability μ_r of:

Given: A material with susceptibility chi = 3 × 10⁻⁴ has a relative permeability μ_r of: These values define the system as per NCERT data. Formula: μ_r = 1 + chi. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: chi = 3 × 10⁻⁴. Substitute: μ_r = 1 + 3 × 10⁻⁴= 1.0003 . 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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CHEMISTRY

What is the magnetic moment (in BM) of [CoClâ‚„]^{2- ?

Given: What is the magnetic moment (in BM) of [CoCl₄]^{2- ? These values define the system as per NCERT data. Formula: Magnetic moment = sqrt3(3+2) = sqrt15 approx 3.87 BM.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Co²⁺ ( d⁷ ) in tetrahedral [CoCl₄]^{2- with weak field Cl^- is high spin ( e⁴ t_2³ ), with 3 unpaired electrons. 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 Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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PHYSICS

An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 ×

Given: An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 8 × 10⁶¹.6 × 10⁻¹⁹ × 0.15 = frac7.28 × 10⁻²⁴².4 × 10⁻²⁰= 3.033 × 10⁻⁴ m approx 0.0303 cm . 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 magnetic dipole experiences a torque of 0.02 N m in a field of 0.4 T at 90° . What is its magnetic moment?

Given: A magnetic dipole experiences a torque of 0.02 N m in a field of 0.4 T at 90° . What is its magnetic moment? These values define the system as per NCERT data. Formula: tau = m B sinθ, so m = tau/B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: tau = 0.02 N m, B = 0.4 T, θ = 90°, sin 90° = 1 . m = 0.02/0.4 × 1 = 0.05 A 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: 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

The magnetic potential energy of a dipole with m = 0.3 A m² in a field B = 0.6 T at 45° is:

Given: The magnetic potential energy of a dipole with m = 0.3 A m² in a field B = 0.6 T at 45° is: These values define the system as per NCERT data. Formula: U_m = -m B cosθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.3 A m², B = 0.6 T, θ = 45°, cos 45° = frac1sqrt2 approx 0.707 . Substitute: U_m = -0.3 × 0.6 × 0.707 approx -0.12726 J approx -0.13 J . 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 dipole with m = 0.35 A m² in B = 0.9 T at 30° has torque:

Given: A dipole with m = 0.35 A m² in B = 0.9 T at 30° has torque: These values define the system as per NCERT data. Formula: tau = m B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.35 A m², B = 0.9 T, θ = 30°, sin 30° = 0.5 . tau = 0.35 × 0.9 × 0.5 = 0.1575 N m . 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 proton moves at 1.5 × 10⁷ m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 1.67 × 1

Given: A proton moves at 1.5 × 10⁷ m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 1.67 × 10⁻²⁷ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac1.67 × 10⁻²⁷ × 1.5 × 10⁷¹.6 × 10⁻¹⁹ × 0.2 = frac2.505 × 10⁻²⁰³.2 × 10⁻²⁰= 0.7828 approx 0.78 m . 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 2.2 m carrying 3.5 A is at 60° to a magnetic field of 0.2 T . What is the force on the wire?

Given: A wire of length 2.2 m carrying 3.5 A is at 60° to a magnetic field of 0.2 T . What is the force on the wire? Formula: Force F = I l B sin θ. Substitution & Calculation: F = 3.5 × 2.2 × 0.2 × sin 60° = 7.7 × 0.2 × 0.866 = 1.3336 approx 1.33 N . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

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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PHYSICS

A bar magnet with original m = 2.0 A m² is cut transversely into two equal parts. What is m of each part?

Given: A bar magnet with original m = 2.0 A m² is cut transversely into two equal parts. What is m of each part? Formula: Given: m = 2.0 A m². Substitution & Calculation: When cut transversely, each part has half the original magnetic moment. . Each part: m' = 2.0/2 = 1.0 A m² . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

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), Chapter: Magnetism and Matter (Latest NCERT 2026-27), Topic: Magnetic dipole moment, bar magnet cut transversely, moment halves m' = m/2. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and illustrative examples. Page.

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PHYSICS

What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment 2 A m² at a distance o

Given: What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment 2 A m² at a distance of 10 cm from its nter? (Take μ_0 = 4π × 10⁻⁷T m A^{-1 ). These values define the system as per NCERT data. Formula: The magnetic field on the equatorial line is B = μ_0/4π m/r³. This is standard NCERT relation. Substitution & Calculation: Given: m = 2 A m², r = 0.1 m, μ_0/4π = 10⁻⁷T m A^{-1 . Substitute: B = 10⁻⁷ × 2/(0.1)³ = 10⁻⁷ × 2/0.001 = 2 × 10⁻⁴T . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Magnetism and Matter, Topic: Bar magnet cut transversely, magnetic moment halves, m' = m/2. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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