An infinitely long wire carrying current I is | vedclass

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Question

(B) The magnetic field due to a finite wire at a perpendicular distance $a$ is given by $B = \frac{\mu_0 I}{4 \pi a} (\sin 	heta_1 + \sin 	heta_2)$.

Vedclass · Accepted Answer

$\frac{{\mu _0}I}{{4\pi a}}\left( {1 - \frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)$

Vedclass · Answer

(B) The magnetic field due to a finite wire at a perpendicular distance $a$ is given by $B = \frac{\mu_0 I}{4 \pi a} (\sin 	heta_1 + \sin 	heta_2)$.Here,the wire extends from $y = b$ to $y = +\infty$.The angle subtended by the upper end at infinity is $	heta_1 = 90^\circ$.The angle subtended by the lower end at $A(0, b)$ is $	heta_2$,where $\sin 	heta_2 = \frac{b}{\sqrt{a^2 + b^2}}$.Since the point $(a, 0)$ is below the end $A$,the angle is taken with a negative sign relative to the perpendicular line from the point to the wire.Thus,$B = \frac{\mu_0 I}{4 \pi a} (\sin 90^\circ - \sin 	heta_2) = \frac{\mu_0 I}{4 \pi a} \left( 1 - \frac{b}{\sqrt{a^2 + b^2}} ight)$.

An infinitely long wire carrying current $I$ is along the $Y$-axis such that its one end is at point $A(0, b)$ while the wire extends up to $+\infty$. The magnitude of the magnetic field strength at point $(a, 0)$ is:

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