An infinitely long wire,located on the z-axis | vedclass

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(A) The magnetic field at a distance $r$ from an infinitely long wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For a path elemen

Vedclass · Accepted Answer

$7 \mu_0 I / 24$

Vedclass · Answer

(A) The magnetic field at a distance $r$ from an infinitely long wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For a path element $d\vec{l}$ along a circular arc of radius $r$,$d\vec{l} = r d	heta$ and $\vec{B}$ is tangential to the path,so $\vec{B} \cdot d\vec{l} = B dl = \frac{\mu_0 I}{2 \pi r} (r d	heta) = \frac{\mu_0 I}{2 \pi} d	heta$.The line integral along the straight path from $(-\sqrt{3} a, a, 0)$ to $(a, a, 0)$ corresponds to an angular sweep. The points lie on the line $y = a$. The distance from the $z$-axis to the line $y=a$ is $r = a / \cos	heta$.However,using the geometry of the path,the angle $	heta$ subtended at the origin changes from $	heta_1$ to $	heta_2$.For point $(-\sqrt{3} a, a, 0)$,$	an	heta_1 = \frac{|-\sqrt{3} a|}{a} = \sqrt{3} \implies 	heta_1 = \frac{\pi}{3}$.For point $(a, a, 0)$,$	an	heta_2 = \frac{a}{a} = 1 \implies 	heta_2 = \frac{\pi}{4}$.The integral is $\int \vec{B} \cdot d\vec{l} = \int_{-	heta_1}^{	heta_2} \frac{\mu_0 I}{2 \pi} d	heta = \frac{\mu_0 I}{2 \pi} (	heta_2 + 	heta_1) = \frac{\mu_0 I}{2 \pi} (\frac{\pi}{4} + \frac{\pi}{3}) = \frac{\mu_0 I}{2 \pi} (\frac{7\pi}{12}) = \frac{7 \mu_0 I}{24}$.

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