A current carrying rectangular loop PQRS is made of uniform wire. The length PR = QS = 5 cm and
PQ = RS = 100 cm. If ammeter current reading changes from I to 2I, the ratio of magnetic forces per unit length on the wire PQ due to wire RS in the two cases respectively ${({f_{PQ}^{I} : f_{PQ}^{2 I}}_{}^{})}_{}^{}$ is:

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1 : 4

1 : 5

1 : 3

1 : 2

answer is D.

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Detailed Solution

Force per unit length between two parallel current carrying wires is,
$F = \frac{μ_{0} i_{1} i_{2}}{2 πr}$
For case(i), $i_{1} = i_{2} = \frac{I}{2}; r = 5 cm$
$\begin{array}{ll} F_{PQ}^{I} = \frac{μ_{0}}{2 π} \frac{I^{2}}{4 \times 5 \times 10^{- 2}} For case(ii) i_{1} = i_{2} = I : r = 5 cm \\ F_{PQ}^{2 I} = \frac{μ_{0}}{2 π} \times \frac{I^{2}}{5 \times 10^{- 2}} F^{I} : F_{Q} \frac{π}{PQ} = \frac{1}{4} : 1 = 1 : 4 \end{array}$