A semi-circular arc of radius r and a straigh | vedclass

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(A) The magnetic field $B$ produced by the semi-circular arc at its center is given by $B = \frac{\mu_{0} i}{4 r}$.The direction of this magnetic fiel

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$\left(\frac{\mu_{0} i^{2}}{4 r}\right)$

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(A) The magnetic field $B$ produced by the semi-circular arc at its center is given by $B = \frac{\mu_{0} i}{4 r}$.The direction of this magnetic field is perpendicular to the plane of the arc (using the right-hand rule).The straight wire carrying current $i$ is placed along the diameter. The small element $P$ at the center of the diameter is in this magnetic field $B$.The magnetic force $dF$ on a small element of length $dl$ carrying current $i$ is given by $dF = i(dl 	imes B)$.Since the magnetic field $B$ is perpendicular to the wire,the magnitude of the force per unit length is $f = \frac{dF}{dl} = iB$.Substituting the value of $B$,we get $f = i \left(\frac{\mu_{0} i}{4 r}ight) = \frac{\mu_{0} i^{2}}{4 r}$.

$A$ semi-circular arc of radius $r$ and a straight wire along the diameter,both are carrying the same current $i$. Find out the magnetic force per unit length on the small element $P$,which is at the center of the diameter.

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