Skip to content
Login Register

#particle physics

10 public questions tagged with this topic.

EM Wave Characteristics - Transverse Nature and E/B Ratio

Gamma rays are produced by which of the following processes?

**EM wave in vacuum** transverse, E and B perpendicular to propagation and to each other, E×B along propagation, in phase, E/B = c =3×10⁸ m/s, c =1/√(μ₀ ε₀), μ₀=4π×10⁻⁷ H/m, ε₀=8.85×10⁻¹² F/m. For E₀=45 V/m, B₀=E₀/c=45/3×10⁸=1.5×10⁻⁷ T=150 nT, illustrating B much smaller than E. Gamma rays are produced in nuclear reactions and emitted by radioactive nuclei, as per the document. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Radioactive decay of the nucleus, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > EM Wave Characteristics - Transverse Nature and E/B Ratio

View Details
Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves with a speed of \( 5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.4 \, \text{T}

**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. Radius r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 5 × 10⁶/1.6 × 10⁻¹⁹ × 0.4) = (8.35 × 10⁻²¹/6.4 × 10⁻²⁰) = 1.3047 × 10⁻¹ m = 13.05 cm . Using F = q v B sinθ,

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

View Details
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

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details
PHYSICS

A proton moves with a speed of 1 × 10⁶ m/s perpendicular to a magnetic field of 0.8 T . What is the radius of its pat

Given: A proton moves with a speed of 1 × 10⁶ m/s perpendicular to a magnetic field of 0.8 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: Radius r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac1.67 × 10⁻²⁷ × 1 × 10⁶¹.6 × 10⁻¹⁹ × 0.8 = frac1.67 × 10⁻²¹¹.28 × 10⁻¹⁹= 1.304 × 10⁻² m = 1.3 cm . Result: The computed value matches the expected outcome and confirms the correct choice. Un

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.

View Details
PHYSICS

What is the energy equivalent of a neutron with mass 1.6749 × 10⁻²⁷kg in Joules? (Given c = 3 × 10⁸m/s )

Given: What is the energy equivalent of a neutron with mass 1.6749 × 10⁻²⁷kg in Joules? (Given c = 3 × 10⁸m/s ) These values define the system as per NCERT data. Formula: E = m c². This is standard NCERT relation. Substitution & Calculation: m = 1.6749 × 10⁻²⁷kg, c² = (3 × 10⁸)² = 9 × 10¹⁶m²/s² . E = 1.6749 × 10⁻²⁷ × 9 × 10¹⁶approx 1.507 × 10⁻¹⁰J . 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,

View Details
PHYSICS

An electron moves with a speed of 4 × 10⁶ m/s perpendicular to a magnetic field of 0.3 T . What is the radius of its

Given: An electron moves with a speed of 4 × 10⁶ m/s perpendicular to a magnetic field of 0.3 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: Radius r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 4 × 10⁶¹.6 × 10⁻¹⁹ × 0.3 = frac3.64 × 10⁻²⁴⁴.8 × 10⁻²⁰= 7.58 × 10⁻⁵ m = 7.58 × 10⁻³ cm . Result: The computed value matches the expected outcome and confirms the correc

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.

View Details