Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

An arrangement of three parallel straight wires placed perpendicular to the plane of paper carry the same current $I$ along the same direction as shown in the figure. The magnitude of force per unit length on the middle wire $B$ is given by:

$\frac{\mu_0 I^2}{2\pi d}$

$\frac{2\mu_0 I^2}{\pi d}$

$\frac{\sqrt{2}\mu_0 I^2}{2\pi d}$

$\frac{\mu_0 I^2}{\sqrt{2}\pi d}$

Step-by-Step Solution

Identify the Physical Law: The magnetic force per unit length between two long, parallel current-carrying conductors separated by a distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi d}$ .
Analyze the Geometry: In the standard configuration for this problem (NEET 2017), wire $B$ is at the right-angle vertex of a right-angled isosceles triangle, with wires $A$ and $C$ at the other two vertices, both at a distance $d$ from $B$ .
Forces on Wire B:

Wire $A$ exerts an attractive force per unit length $f_1 = \frac{\mu_0 I^2}{2\pi d}$ on $B$ (since the currents are in the same direction) .
Wire $C$ exerts an attractive force per unit length $f_2 = \frac{\mu_0 I^2}{2\pi d}$ on $B$ .

Vector Summation: Since the wires are arranged such that $AB$ is perpendicular to $BC$ , the force vectors $\mathbf{f_1}$ and $\mathbf{f_2}$ are mutually perpendicular. The magnitude of the resultant force per unit length is: $F_{net} = \sqrt{f_1^2 + f_2^2} = \sqrt{2} \times f = \sqrt{2} \frac{\mu_0 I^2}{2\pi d}$
Simplification: $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ , so the magnitude is $\frac{\mu_0 I^2}{\sqrt{2}\pi d}$ .

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