(A) The force per unit length between two parallel current-carrying conductors is given by the formula:$f = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r}$Giv

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$1.6 \times 10^{-5}\,N$ (attractive)

(A) The force per unit length between two parallel current-carrying conductors is given by the formula:$f = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r}$Given:$I_1 = 5\,A$,$I_2 = 8\,A$,$r = 0.5\,m$,and $\frac{\mu_0}{4\pi} = 10^{-7}\,T\cdot m/A$.Substituting the values:$f = 10^{-7} \times \frac{2 \times 5 \times 8}{0.5}$$f = 10^{-7} \times \frac{80}{0.5}$$f = 10^{-7} \times 160$$f = 1.6 \times 10^{-5}\,N/m$.Since the currents are in the same direction,the force is attractive.

Class 12 Physics Moving Charges and Magnetism Force on a Current Carrying Conductor

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Two long straight parallel conductors separated by a distance of $0.5\,m$ carry currents of $5\,A$ and $8\,A$ in the same direction. The force per unit length experienced by each other is

A
$1.6 \times 10^{-5}\,N$ (attractive)
B
$1.6 \times 10^{-5}\,N$ (repulsive)
C
$16 \times 10^{-5}\,N$ (attractive)
D
$16 \times 10^{-5}\,N$ (repulsive)

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

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Two long straight parallel conductors separated by a distance of $0.5\,m$ carry currents of $5\,A$ and $8\,A$ in the same direction. The force per unit length experienced by each other is

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