A current of 8 text{ A} each flows in opposit | vedclass

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(B) Let the two wires be parallel and separated by a distance $d = 30 	ext{ cm} = 0.3 	ext{ m}$.The current in each wire is $I = 8 	ext{ A}$.Since

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$300$

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(B) Let the two wires be parallel and separated by a distance $d = 30 	ext{ cm} = 0.3 	ext{ m}$.The current in each wire is $I = 8 	ext{ A}$.Since the currents flow in opposite directions, the magnetic fields produced by both wires at the midpoint (distance $r = d/2 = 0.15 	ext{ m}$) will point in the same direction according to the Right-Hand Thumb Rule.The magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$.For the first wire, $B_1 = \frac{\mu_0 I}{2\pi (d/2)} = \frac{2 	imes 10^{-7} 	imes 8}{0.15} = \frac{16 	imes 10^{-7}}{0.15} = \frac{1600}{15} 	imes 10^{-7} \approx 106.67 \mu 	ext{T}$.Since both wires contribute equally in the same direction, the total magnetic field $B_{total} = B_1 + B_2 = 2 	imes B_1 = 2 	imes 106.67 \mu 	ext{T} = 213.33 \mu 	ext{T}$.Given the options provided, $213.33 \mu 	ext{T}$ is closest to $200 \mu 	ext{T}$, but based on standard physics problem sets where $d$ is often adjusted to yield integer results (e.g.,if $d = 0.2 	ext{ m}$), $213.33 \mu 	ext{T}$ is the calculated value. Among the choices, $213.33$ is not present, but $213.33$ is mathematically correct.

$A$ current of $8 \text{ A}$ each flows in opposite directions in two parallel conducting wires placed at a distance of $30 \text{ cm}$. The magnitude of the magnetic field at the midpoint between the two wires is . . . . . . $\mu \text{T}$. (Given: $\frac{\mu_0}{4\pi} = 10^{-7} \text{ N/A}^2$)

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