A straight wire carrying a current of $12\; A$ is bent into a semi-circular arc of radius $2.0\; cm$ as shown in Figure $(a)$. Consider the magnetic field $B$ at the centre of the arc.
$(a)$ What is the magnetic field due to the straight segments?
$(b)$ In what way the contribution to $B$ from the semicircle differs from that of a circular loop and in what way does it resemble?
$(c)$ Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Figure $(b)$
A straight wire carrying a current of $12\; A$ is bent into a semi-circular arc of radius $2.0\; cm$ as shown in Figure $(a)$. Consider the magnetic field $B$ at the centre of the arc.
$(a)$ What is the magnetic field due to the straight segments?
$(b)$ In what way the contribution to $B$ from the semicircle differs from that of a circular loop and in what way does it resemble?
$(c)$ Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Figure $(b)$
$(a)$ $dl$ and $r$ for each element of the straight segments are parallel. Therefore, $d l \times r =0 .$ Straight segments do not contribute to $| B |$
$(b)$ For all segments of the semicircular arc, $d l \times r$ are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of $B$ for a semicircular arc
is given by the right-hand rule and magnitude is half that of a circular loop. Thus $B$ is $1.9 \times 10^{-4} T$ normal to the plane of the paper going into it.
$(c)$ Same magnitude of $B$ but opposite in direction to that in $(b).$
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- [AIPMT 1997]
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