Force on Current-Carrying Conductor and Between Parallel Wires

Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.07 \, \text{m} \) apart carry \( 8 \, \text{A} \) and \( 6 \, \text{A} \) in the same direction.

**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 × 8 × 6/2 π × 0.07) = (192 × 10⁻⁷/0.14) = 1.3714 × 10⁻⁵ ≈ 1.37 × 10⁻⁵ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B =

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.11 \, \text{m} \) apart carry currents of \( 6 \, \text{A} \) and \( 2 \, \text{A} \) in the sam

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 6 × 2/2 π × 0.11) = (48 × 10⁻⁷/0.22) = 2.18 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.07 \, \text{m} \) apart carry currents of \( 9 \, \text{A} \) and \( 2 \, \text{A} \) in the sam

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 9 × 2/2 π × 0.07) = (72 × 10⁻⁷/0.14) = 5.14 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 2.3 \, \text{m} \) carrying \( 4 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.6 \

**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 = I l B sin θ . F = 4 × 2.3 × 0.6 × sin 60° = 5.52 × 0.866 = 4.7803 ≈ 4.78 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.12 \, \text{m} \) apart carry \( 7 \, \text{A} \) and \( 3 \, \text{A} \) in the same direction.

**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 × 3/2 π × 0.12) = (84 × 10⁻⁷/0.24) = 3.5 × 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)

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.06 \, \text{m} \) apart carry \( 7 \, \text{A} \) and \( 5 \, \text{A} \) in the same direction.

**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 × 7 × 5/2 π × 0.06) = (140 × 10⁻⁷/0.12) = 1.166 × 10⁻⁵ ≈ 1.17 × 10⁻⁵ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B =

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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

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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

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 0.8 \, \text{m} \) carrying \( 3 \, \text{A} \) makes an angle of \( 30^\circ \) with a magnetic fie

**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. Force F = I l B sin θ . F = 3 × 0.8 × 0.6 × sin 30° = 2.4 × 0.6 × 0.5 = 0.72 N . 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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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 1.6 \, \text{m} \) carrying \( 5 \, \text{A} \) is at \( 45^\circ \) to a magnetic field of \( 0.1 \

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. Force F = I l B sin θ . F = 5 × 1.6 × 0.1 × sin 45° = 8 × 0.1 × 0.707 = 0.5656 ≈ 0.57 N . 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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Force on Current-Carrying Conductor and Between Parallel Wires

A straight wire of length \( 0.9 \, \text{m} \) carries a current of \( 7 \, \text{A} \) perpendicular to a uniform magn

**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. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 7 × 0.9 × 0.2 = 1.26 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.03 \, \text{m} \) apart carry currents of \( 10 \, \text{A} \) and \( 2 \, \text{A} \) in the sa

**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. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 10 × 2/2 π × 0.03) = (80 × 10⁻⁷/0.06) = 1.333 × 10⁻⁵ ≈ 1.33 × 10⁻⁵ N/m . Using F = q v B sinθ, F = I l B sinθ, B =

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