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Q.

Two long parallel wires are kept at a separation $r$ . Magnitude of magnetic force experienced by unit length of each wire is F. If the separation between the wire is halved and current in each wire is doubled, then magnitude of force experienced by each wire will be

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a

F/4

b

2F

c

8F

d

4F

answer is C.

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Detailed Solution

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$\begin{array}{l} F = \frac{μ_{0} I^{2}}{2 π r}, F_{1} = \frac{μ_{0} {(I_{1})}^{2}}{2 π r_{1}} \\ \Rightarrow \frac{F_{1}}{F} = {(\frac{I_{1}}{I})}^{2} . (\frac{r}{r_{1}}) = {(\frac{2 I}{I})}^{2} . (\frac{r}{r / 2}) = 8 \\ \Rightarrow F_{1} = 8 F \end{array}$

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