A circular arc of radius r carrying current I | vedclass

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(C) The magnetic field $B$ at the centre of a circular arc carrying current $I$ is given by the formula: $B = \frac{\mu_0 I}{2r} \left( \frac{	heta}{

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$\frac{\mu_0 I}{64 r}$

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(C) The magnetic field $B$ at the centre of a circular arc carrying current $I$ is given by the formula: $B = \frac{\mu_0 I}{2r} \left( \frac{	heta}{2\pi} ight)$.Given that the angle subtended at the centre is $	heta = \frac{\pi}{16}$.Substituting the value of $	heta$ into the formula:$B = \frac{\mu_0 I}{2r} \left( \frac{\pi/16}{2\pi} ight)$.Simplifying the expression inside the parentheses:$\frac{\pi/16}{2\pi} = \frac{\pi}{16 	imes 2\pi} = \frac{1}{32}$.Now,substitute this back into the equation for $B$:$B = \frac{\mu_0 I}{2r} 	imes \frac{1}{32} = \frac{\mu_0 I}{64r}$.

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{16}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is [where $\mu_0$ is the permeability of free space].

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