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(C) The magnetic field $B$ at a point $P$ due to a finite straight wire carrying current $i$ is given by the formula:$B = \frac{{{\mu _0}i}}{{4\pi r}}

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$\frac{{\sqrt 2 {\mu _0}i}}{{8\pi l}}$

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(C) The magnetic field $B$ at a point $P$ due to a finite straight wire carrying current $i$ is given by the formula:$B = \frac{{{\mu _0}i}}{{4\pi r}}(\sin {\phi _1} + \sin {\phi _2})$where $r$ is the perpendicular distance from the wire to the point $P$,and $\phi_1, \phi_2$ are the angles subtended by the ends of the wire at point $P$.In this case,the perpendicular distance $r = l$. The angle subtended by the top end is $\phi_1 = 0^\circ$. The angle subtended by the bottom end is $\phi_2 = 45^\circ$ (since the wire length is $l$ and distance is $l$,$	an \phi_2 = l/l = 1$,so $\phi_2 = 45^\circ$).Substituting these values into the formula:$B = \frac{{{\mu _0}i}}{{4\pi l}}(\sin 0^\circ + \sin 45^\circ)$$B = \frac{{{\mu _0}i}}{{4\pi l}}(0 + \frac{1}{{\sqrt 2 }})$$B = \frac{{{\mu _0}i}}{{4\sqrt 2 \pi l}} = \frac{{\sqrt 2 {\mu _0}i}}{{8\pi l}}$

The figure shows a straight wire of length $l$ carrying current $i$. The magnitude of the magnetic field produced by the current at point $P$ is:

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