Two long and parallel straight wires A and B | vedclass

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(A) Current in wire $A$,$I_{A} = 8.0 \, A$.Current in wire $B$,$I_{B} = 5.0 \, A$.Distance between the wires,$r = 4.0 \, cm = 0.04 \, m$.Length of the

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$2 	imes 10^{-5} \, N$ Attractive

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(A) Current in wire $A$,$I_{A} = 8.0 \, A$.Current in wire $B$,$I_{B} = 5.0 \, A$.Distance between the wires,$r = 4.0 \, cm = 0.04 \, m$.Length of the section of wire $A$,$l = 10 \, cm = 0.1 \, m$.The force per unit length between two parallel wires is given by $f = \frac{\mu_{0} I_{A} I_{B}}{2 \pi r}$.The total force $F$ on a length $l$ is $F = \frac{\mu_{0} I_{A} I_{B} l}{2 \pi r}$.Substituting the values:$F = \frac{(4 \pi 	imes 10^{-7} \, T \cdot m/A) 	imes 8.0 \, A 	imes 5.0 \, A 	imes 0.1 \, m}{2 \pi 	imes 0.04 \, m}$.$F = \frac{2 	imes 10^{-7} 	imes 40 	imes 0.1}{0.04} \, N$.$F = \frac{8 	imes 10^{-7}}{0.04} \, N = 2 	imes 10^{-5} \, N$.Since the currents are in the same direction,the force is attractive.

Two long and parallel straight wires $A$ and $B$ carrying currents of $8.0 \, A$ and $5.0 \, A$ in the same direction are separated by a distance of $4.0 \, cm$. Estimate the force on a $10 \, cm$ section of wire $A$.

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