A current I=5 text{ A} flows along a thin wir | vedclass

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(A) The magnetic field at the center $O$ is the vector sum of the fields due to the curved part and the straight part.$1$. Magnetic field due to the c

Vedclass · Accepted Answer

$33.6$

Vedclass · Answer

(A) The magnetic field at the center $O$ is the vector sum of the fields due to the curved part and the straight part.$1$. Magnetic field due to the curved part:The angle subtended by the arc at the center is $	heta = 2\pi - 2\phi = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.$B_{	ext{arc}} = \frac{\mu_0 I 	heta}{4\pi R} = \frac{10^{-7} 	imes 5 	imes (3\pi/2)}{0.1} = 50 	imes 10^{-7} 	imes 1.5 	imes 3.14 \approx 23.55 \mu	ext{T}$.$2$. Magnetic field due to the straight part:The distance $d$ from $O$ to the straight wire is $d = R \cos(45^{\circ}) = R/\sqrt{2}$.The angles subtended by the ends of the wire at $O$ are $\phi_1 = 45^{\circ}$ and $\phi_2 = 45^{\circ}$.$B_{	ext{straight}} = \frac{\mu_0 I}{4\pi d} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{10^{-7} 	imes 5}{0.1/\sqrt{2}} 	imes (1/\sqrt{2} + 1/\sqrt{2}) = \frac{10^{-7} 	imes 5 	imes \sqrt{2}}{0.1} 	imes \frac{2}{\sqrt{2}} = \frac{10^{-6} 	imes 5 	imes 2}{0.1} = 10 \mu	ext{T}$.$3$. Total magnetic field:Since both fields point in the same direction (into the page), $B_{	ext{net}} = B_{	ext{arc}} + B_{	ext{straight}} = 23.55 \mu	ext{T} + 10 \mu	ext{T} = 33.55 \mu	ext{T} \approx 33.6 \mu	ext{T}$.

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