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Question

(A) The magnetic field $B$ due to a finite straight wire at a perpendicular distance $a$ from its centre is given by $B = \frac{{\mu _0}i}{4\pi a}(\si

Vedclass · Accepted Answer

$\frac{4{\mu _0}i}{\sqrt 5 \pi L}$

Vedclass · Answer

(A) The magnetic field $B$ due to a finite straight wire at a perpendicular distance $a$ from its centre is given by $B = \frac{{\mu _0}i}{4\pi a}(\sin {\phi _1} + \sin {\phi _2})$.Here,$a = \frac{L}{4}$ and $\phi _1 = \phi _2 = \phi$.From the geometry,$\sin \phi = \frac{L/2}{\sqrt{(L/2)^2 + (L/4)^2}} = \frac{L/2}{\sqrt{L^2/4 + L^2/16}} = \frac{L/2}{\sqrt{5L^2/16}} = \frac{L/2}{\sqrt{5}L/4} = \frac{2}{\sqrt{5}}$.Substituting these values into the formula:$B = \frac{{\mu _0}i}{4\pi (L/4)} 	imes (2 \sin \phi) = \frac{{\mu _0}i}{\pi L} 	imes 2 	imes \frac{2}{\sqrt{5}} = \frac{4{\mu _0}i}{\sqrt{5}\pi L}$.

$A$ current $i$ is flowing in a straight conductor of length $L$. The magnetic induction at a point distant $\frac{L}{4}$ from its centre will be

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