Two long straight wires P and Q carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F 1 . If distan

Question

Two long straight wires P and Q carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F 1 . If distance between wires is halved and currents on them are doubled, force F 2 on 10 cm length of wire P will be:

TestHub · Accepted Answer

As we know, force per unit length between two parallel straight wires is given by, $F = \frac{μ_{0} i_{1} i_{2}}{2 π d}$ .

Therefore, required ratio

$\frac{F_{1}}{F_{2}} = \frac{\frac{μ_{0} (10)^{2}}{2 π (5 \times 10^{- 2})}}{\frac{μ_{0} (20)^{2}}{2 π (\frac{5}{2} \times 10^{- 2})}} = \frac{1}{8}$

$\Rightarrow F_{2} = 8 F_{1}$

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