Magnetic Force on Moving Charge - Lorentz Force and Motion

Magnetic Force on Moving Charge - Lorentz Force and Motion

Two parallel wires \( 0.1 \, \text{m} \) apart carry \( 8 \, \text{A} \) and \( 6 \, \text{A} \) in the same direction.

**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. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 8 × 6/2 π × 0.1) = (192 × 10⁻⁷/0.2) = 9.6 × 10⁻⁵ N/m . Using F = q v B sinθ, F =

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Magnetic Force on Moving Charge - Lorentz Force and Motion

Two parallel wires \( 0.01 \, \text{m} \) apart carry \( 5 \, \text{A} \) and \( 6 \, \text{A} \) in opposite directions

**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. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 5 × 6/2 π × 0.01) = (120 × 10⁻⁷/0.02) = 6 × 10⁻⁵ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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Magnetic Force on Moving Charge - Lorentz Force and Motion

Two parallel wires \( 0.06 \, \text{m} \) apart carry \( 9 \, \text{A} \) and \( 5 \, \text{A} \) in the same direction.

**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. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 9 × 5/2 π × 0.06) = (180 × 10⁻⁷/0.12) = 1.5 × 10⁻⁵ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

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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

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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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.07 \, \text{m}^2 \) with 10 turns carries \( 5 \, \text{A} \) in a field of \( 0.8 \, \t

**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. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 10 × 5 × 0.07 × 0.8 × 1 = 2.8 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.09 \, \text{m}^2 \) with 18 turns carries \( 2 \, \text{A} \) in a field of \( 0.6 \, \t

**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. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 18 × 2 × 0.09 × 0.6 × 1 = 1.944 ≈ 1.94 N m . Using F = q v B sinθ, F = I l B sinθ, B =

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.03 \, \text{m}^2 \) with 20 turns carries \( 4 \, \text{A} \) in a field of \( 0.9 \, \t

**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. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 20 × 4 × 0.03 × 0.9 × 1 = 2.16 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

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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

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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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

What is the direction of the magnetic force on a positive charge moving parallel to a uniform magnetic field?

**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. The magnetic force is given by F = q (v × B) . If the velocity v is parallel to the magnetic field B , the angle between them is 0° , so sin 0° = 0 , and the force magnitude is zero. Thus, there is no force. Using F = q v B sinθ, F

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A long wire carries \( 18 \, \text{A} \). At what distance is the magnetic field \( 6 \times 10^{-6} \, \text{T} \)? (\(

**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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 18/2 π × 6 × 10⁻⁶) = (72 × 10⁻⁷/12 × 10⁻⁶) = 0.6 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B

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