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#proton

12 public questions tagged with this topic.

Solenoid, Toroid and Ampere's Law

A proton moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a uniform magnetic field of \( 0.5 \,

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Radius r = (mv/qB) . Substitute: r = (1.67 × 10⁻²⁷ × 2 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.34 × 10⁻²¹/8 × 10⁻²⁰) = 4.175 × 10⁻² m = 4.18 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A proton moves at \( 4 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.05 \, \text{T} \). What is the radi

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 4 × 10⁷/1.6 × 10⁻¹⁹ × 0.05) = (6.68 × 10⁻²⁰/8 × 10⁻²¹) = 8.35 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A B

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 3.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.08 \, \text{T} \). What is the ra

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 3.5 × 10⁷/1.6 × 10⁻¹⁹ × 0.08) = (5.845 × 10⁻²⁰/1.28 × 10⁻²⁰) = 4.5664 ≈ 4.57 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 2.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.15 \, \text{T} \). What is the ma

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 2.5 × 10⁷ × 0.15 = 6 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 7.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.12 \, \text{T} \). What is the ma

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 7.5 × 10⁷ × 0.12 = 1.44 × 10⁻¹² N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves with a speed of \( 1.8 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.5 \, \text{

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. Radius r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 1.8 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.006 × 10⁻²¹/8 × 10⁻²⁰) = 3.7575 × 10⁻² m = 3.76 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 6 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.1 \, \text{T} \). What is the magne

**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. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 6 × 10⁷ × 0.1 = 9.6 × 10⁻¹³ N . Using F = q

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 3 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.1 \, \text{T} \). What is the radiu

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 3 × 10⁷/1.6 × 10⁻¹⁹ × 0.1) = (5.01 × 10⁻²⁰/1.6 × 10⁻²⁰) = 3.13125 m ≈ 3.13 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

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

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