An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current 'I' along the same direction as shown in figure. Magnitude of force per unit length on the middle wire 'B' is given by (a) (2μ₀ I²)/(π d) (b) (√(2)μ₀ I²)/(π d) (c) (μ₀ I²)/(√(2)π d) (d) (μ₀ I²)/(2π d)

Answer: (c). BCAff√2 f Parallel currents in the same direction attract. Force per unit length between two wires: f = (μ₀ I²)/(2π d). On B: fA points from B towards A and fC points from B towards C, each (μ₀ I²)/(2π d), at 90° to each other. fnet = √(2) (μ₀ I²)/(2π d) = (μ₀ I²)/(√(2) π d) So the answer is (c) (μ₀ I²)/(√(2)π d).

An arrangement of three parallel straight wires placed perpendicular… | NEET 2017

Q51NEET 2017Medium

Force on Current-Carrying Conductors

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current $\text{'}I\text{'}$ along the same direction as shown in figure. Magnitude of force per unit length on the middle wire $\text{'}B\text{'}$ is given by

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Answer: (c)

Parallel currents in the same direction attract. Force per unit length between two wires: $f = \frac{\mu_0 I^2}{2\pi d}$ .
On B: $f_A$ points from B towards A and $f_C$ points from B towards C, each $\frac{\mu_0 I^2}{2\pi d}$ , at $90^\circ$ to each other.
$f_{net} = \sqrt{2}\,\frac{\mu_0 I^2}{2\pi d} = \frac{\mu_0 I^2}{\sqrt{2}\,\pi d}$
So the answer is (c) $\frac{\mu_0 I^2}{\sqrt{2}\pi d}$ .

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