#opposite currents

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)

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

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Force on Current-Carrying Conductor and Between Parallel Wires

What is the nature of the force between two parallel wires carrying currents in opposite directions?

**Parallel current interaction** arises because each wire's field B = μ₀ I/(2π d) exerts force F = I l B on other. Force per length f = B I, leading to f = μ₀ I₁ I₂/(2π d). This defines ampere: two wires 1 m apart carrying 1 A exert 2×10⁻⁷ N/m. Two parallel wires with currents in opposite directions experience a repulsive force because the magnetic field produced by one wire interacts with the current in the other, causing them to push apart. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.09 \, \text{m} \) apart carry \( 4 \, \text{A} \) and \( 5 \, \text{A} \) in opposite directions

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 4 × 5/2 π × 0.09) = (80 × 10⁻⁷/0.18) = 4.44 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.08 \, \text{m} \) apart carry \( 7 \, \text{A} \) and \( 4 \, \text{A} \) in opposite directions

**Parallel current interaction** arises because each wire's field B = μ₀ I/(2π d) exerts force F = I l B on other. Force per length f = B I, leading to f = μ₀ I₁ I₂/(2π d). This defines ampere: two wires 1 m apart carrying 1 A exert 2×10⁻⁷ N/m. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 7 × 4/2 π × 0.08) = (112 × 10⁻⁷/0.16) = 7 × 10⁻⁶ N/m . 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 > Force on Current-Carrying Conductor and Between Parallel Wires

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.04 \, \text{m} \) apart carry \( 5 \, \text{A} \) and \( 7 \, \text{A} \) in opposite directions

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 5 × 7/2 π × 0.04) = (140 × 10⁻⁷/0.08) = 1.75 × 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

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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