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#Electromagnetism

102 public questions tagged with this topic.

PHYSICS

An electron moves at 6.5 × 10⁶m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 9.1 × 10⁻³

Given: An electron moves at 6.5 × 10⁶m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹kg, charge = 1.6 × 10⁻¹⁹C ) Formula: r = mv/qB. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 6.5 × 10⁶¹.6 × 10⁻¹⁹ × 0.2 = frac5.915 × 10⁻²⁴³.2 × 10⁻²⁰= 1.848 × 10⁻⁴m approx 0.0185 cm . 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 circular loop of radius 18 cm is deformed into a straight wire in a 0.14 T field in 0.7 s. What is the induced emf?

Given: A circular loop of radius 18 cm is deformed into a straight wire in a 0.14 T field in 0.7 s. What is the induced emf? These values define the system as per NCERT data. Formula: Initial flux: Phi = B A = 0.14 × π × (0.18)² = 0.01425 Wb. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Final flux = 0. varepsilon = Δ Phi/Δ t = 0.01425/0.7 = 0.02036 V approx 0.02 V . 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: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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PHYSICS

A solenoid of 900 turns/m and area 0.015 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m )

Given: A solenoid of 900 turns/m and area 0.015 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m ) These values define the system as per NCERT data. Formula: L = μ_r μ_0 n² A l, assume l = 1 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: L = 2 × 4π × 10⁻⁷ × (900)² × 0.015 × 1 = 0.0305 H approx 0.03 H . 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 solenoid of 1000 turns/m and area 0.02 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m )

Given: A solenoid of 1000 turns/m and area 0.02 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m ) These values define the system as per NCERT data. Formula: L = μ_r μ_0 n² A l, assume l = 1 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: L = 2 × 4π × 10⁻⁷ × (1000)² × 0.02 × 1 = 0.05024 H approx 0.05 H . 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 rectangular loop of sides 12 cm and 4 cm moves out of a 0.35 T field at 2 m/s perpendicular to the longer side. What i

Given: A rectangular loop of sides 12 cm and 4 cm moves out of a 0.35 T field at 2 m/s perpendicular to the longer side. What is the motional emf? These values define the system as per NCERT data. Formula: varepsilon = B l v, l = 0.04 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 0.35 × 0.04 × 2 = 0.028 V . 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: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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PHYSICS

A solenoid of 250 turns and length 0.8 m induces an emf of 0.75 V in a nearby coil when its current changes from 0 to 2.

Given: A solenoid of 250 turns and length 0.8 m induces an emf of 0.75 V in a nearby coil when its current changes from 0 to 2.5 A in 0.25 s. What is the mutual inductance? These values define the system as per NCERT data. Formula: varepsilon = M Δ I/Δ t. This is standard NCERT relation. Substitution & Calculation: Δ I = 2.5 - 0 = 2.5 A, Δ t = 0.25 s . M = fracvarepsilonΔ I/Δ t = frac0.752.5/0.25 = 0.75/10 = 0.075 H . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A square loop of side 30 cm rotates at 12 rad/s in a 0.1 T field. What is the maximum emf induced?

Given: A square loop of side 30 cm rotates at 12 rad/s in a 0.1 T field. What is the maximum emf induced? These values define the system as per NCERT data. Formula: A = (0.3)² = 0.09 m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon_0 = N B A omega = 1 × 0.1 × 0.09 × 12 = 0.108 V . 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: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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PHYSICS

The magnetic potential energy of a dipole with m = 0.6 A m² in a field B = 0.2 T at 0° is:

Given: The magnetic potential energy of a dipole with m = 0.6 A m² in a field B = 0.2 T at 0° 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 = 0.6 A m², B = 0.2 T, θ = 0°, cos 0° = 1 . Substitute: U_m = -0.6 × 0.2 × 1 = -0.12 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 rod of length 0.2 m moves at 5 m/s in a 0.3 T field perpendicular to its length. What is the induced emf?

Given: A rod of length 0.2 m moves at 5 m/s in a 0.3 T field perpendicular to its length. What is the induced emf? These values define the system as per NCERT data. Formula: Motional emf: varepsilon = B l v. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 0.3 × 0.2 × 5 = 0.3 V . 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: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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PHYSICS

A square loop of side 26 cm rotates at 12 rad/s in a 0.3 T field. What is the maximum emf induced?

Given: A square loop of side 26 cm rotates at 12 rad/s in a 0.3 T field. What is the maximum emf induced? Formula: A = (0.26)² = 0.0676 m². Substitution & Calculation: ε_0 = N B A omega = 1 × 0.3 × 0.0676 × 12 = 0.24336 V approx 0.243 V . 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: Electromagnetic Induction (Latest NCERT 2026-27), Topic: Rotating coil, maximum emf ε₀ = NBAω, N = 300, B = 0.07 T, A = 0.012 m². The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI.

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PHYSICS

A rod rotates at 15 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.4 m, what is the emf induced?

Given: A rod rotates at 15 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.4 m, what is the emf induced? These values define the system as per NCERT data. Formula: varepsilon = 1/2 B omega R². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 1/2 × 0.3 × 15 × (0.4)² = 0.36 V . 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: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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PHYSICS

A magnetic dipole of moment 0.6 A m² is in a uniform field of 0.4 T at 60° . What is the torque on it?

Given: A magnetic dipole of moment 0.6 A m² is in a uniform field of 0.4 T at 60° . What is the torque on it? These values define the system as per NCERT data. Formula: Torque is tau = m B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.6 A m², B = 0.4 T, θ = 60°, sin 60° = fracsqrt32 approx 0.866 . Substitute: tau = 0.6 × 0.4 × 0.866 approx 0.20784 N m approx 0.21 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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