Two long straight wires and carrying equal current 10 A each were kept parallel to each other at 5 cm distance. [2023]

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Solution

Q.
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 the distance between the wires is halved and the currents on them are doubled, the force $F_{2}$ on 10 cm length of wire $P$ will be [2023]

1 $10 F_{1}$

2 $8 F_{1}$

3 $\frac{F_{1}}{8}$

4 $\frac{F_{1}}{10}$

Ans.
(2)

$Force per unit length between two parallel straight wires$

$= \frac{μ_{0} i_{1} i_{2}}{2 π d}$

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

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