Two parallel wires \( 0.04 \, \text{m} \) apart carry \( 10 \, \text{A} \) and \( 4 \, \text{A} \) in opposite 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 × 10 × 4/2 π × 0.04) = (160 × 10⁻⁷/0.08) = 2 × 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