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#current wire

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Force on Current-Carrying Conductor and Between Parallel Wires

A straight wire of length \( 1.1 \, \text{m} \) carries a current of \( 4 \, \text{A} \) perpendicular to a uniform magn

**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 , where θ = 90° , so sin θ = 1 . F = 4 × 1.1 × 0.3 = 1.32 N . 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

A wire of length \( 1.9 \, \text{m} \) carrying \( 4.5 \, \text{A} \) is at \( 45^\circ \) to a magnetic field of \( 0.3

**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 = 4.5 × 1.9 × 0.3 × sin 45° = 8.55 × 0.3 × 0.707 = 1.8127 ≈ 1.81 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

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

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