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(C) Let the wires be at $A$ and $B$ with currents $I_1 = 8 	ext{ A}$ and $I_2 = 15 	ext{ A}$ in opposite directions. The distance $AB = 7 	ext{ cm}

Vedclass · Accepted Answer

$68$

Vedclass · Answer

(C) Let the wires be at $A$ and $B$ with currents $I_1 = 8 	ext{ A}$ and $I_2 = 15 	ext{ A}$ in opposite directions. The distance $AB = 7 	ext{ cm}$.Point $P$ is equidistant from $A$ and $B$, so $PA = PB = r$. Since $\angle APB = 90^\circ$, in $	riangle APB$, $AB^2 = PA^2 + PB^2 = 2r^2$.$7^2 = 2r^2 \implies 49 = 2r^2 \implies r^2 = 24.5 \implies r = \sqrt{24.5} 	ext{ cm} = \sqrt{24.5} 	imes 10^{-2} 	ext{ m}$.The magnetic field due to wire $A$ is $B_1 = \frac{\mu_0 I_1}{2\pi r}$ and due to wire $B$ is $B_2 = \frac{\mu_0 I_2}{2\pi r}$.Since the currents are in opposite directions, the fields $B_1$ and $B_2$ at $P$ are perpendicular to each other.The resultant magnetic field $B = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{2\pi r} \sqrt{I_1^2 + I_2^2}$.$B = \frac{2 	imes 10^{-7}}{r} \sqrt{8^2 + 15^2} = \frac{2 	imes 10^{-7}}{\sqrt{24.5} 	imes 10^{-2}} 	imes 17 = \frac{34 	imes 10^{-5}}{\sqrt{24.5}} \approx \frac{34 	imes 10^{-5}}{4.95} \approx 6.86 	imes 10^{-5} 	ext{ T}$.Given $B = X 	imes 10^{-6} 	ext{ T}$, so $X \approx 68.6 \approx 68$.

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