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(B) The magnetic field at point $P$ is produced by the two semi-infinite segments of the wire.Using the formula for the magnetic field due to a semi-i

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$10^{-4} \, N/m$

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(B) The magnetic field at point $P$ is produced by the two semi-infinite segments of the wire.Using the formula for the magnetic field due to a semi-infinite wire,$B = \frac{\mu_0 i}{4 \pi r}$.At point $P$,the distance $r = 5 \, cm = 5 	imes 10^{-2} \, m$.The total magnetic field $B$ at $P$ is the sum of the fields from the two semi-infinite segments:$B = \frac{\mu_0 i}{4 \pi r} + \frac{\mu_0 i}{4 \pi r} = 2 	imes \frac{\mu_0 i}{4 \pi r}$Substituting the values: $B = 2 	imes \frac{10^{-7} 	imes 5}{5 	imes 10^{-2}} = 2 	imes 10^{-5} \, T$.The force per unit length on a current-carrying wire is given by $f = iB$.$f = 5 \, A 	imes 2 	imes 10^{-5} \, T = 10^{-4} \, N/m$.

Find the force per unit length at point $P$. The current in the wire is $5 \, A$ and the distance between the parallel segments is $10 \, cm$.

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