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Q.

Figure shows three long straight wires parallel and equally spaced with identical space. Then, magnitude of force acting on each wire due to the other is
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a

(a) $F_{a} > F_{b} > F_{c}$

b

(b) $F_{b} > F_{c} > F_{a}$

c

(c) $F_{c} > F_{a} > F_{b}$

d

(d) $F_{b} > F_{a} > F_{c}$

answer is B.

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Detailed Solution

$F_{a} = F_{b a} - F_{c a} = \frac{μ_{0}}{2 π} \frac{I^{2}}{r} - \frac{μ_{0}}{2 π} \frac{I^{2}}{2 r} = \frac{μ_{0}}{4 π} \frac{I^{2}}{r} F_{b} = F_{a b} + F_{c b} = \frac{μ_{0}}{2 π} \frac{I^{2}}{r} + \frac{μ_{0}}{2 π} \frac{I^{2}}{r} = \frac{μ_{0}}{π} \frac{I^{2}}{r} F_{c} = F_{a c} + F_{b c} = \frac{μ_{0}}{2 π} \frac{I^{2}}{2 r} + \frac{μ_{0}}{2 π} \frac{I^{2}}{r} = \frac{3 μ_{0}}{4 π} \frac{I^{2}}{r} ∴ F_{b} > F_{c} > F_{a}$

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