A very long conducting wire is bent in a semi | vedclass

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(D) The magnetic field at point $P$ is the sum of the fields due to the two semi-infinite straight wires and the semicircular arc.$1$. The magnetic fi

Vedclass · Accepted Answer

$\frac{\mu_0 i }{4 R }\left[1-\frac{2}{\pi}\right]$ pointed away from the page

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(D) The magnetic field at point $P$ is the sum of the fields due to the two semi-infinite straight wires and the semicircular arc.$1$. The magnetic field due to each semi-infinite wire at distance $R$ is $B_{straight} = \frac{\mu_0 i}{4 \pi R}$. Using the right-hand rule,both wires produce a magnetic field pointing into the page.$2$. The magnetic field due to the semicircular arc of radius $R$ is $B_{arc} = \frac{\mu_0 i}{4 R}$. Using the right-hand rule,this field points out of the page.$3$. The net magnetic field is $B_{net} = B_{arc} - 2 	imes B_{straight} = \frac{\mu_0 i}{4 R} - 2 \left( \frac{\mu_0 i}{4 \pi R} ight) = \frac{\mu_0 i}{4 R} \left( 1 - \frac{2}{\pi} ight)$.Since $1 > \frac{2}{\pi}$,the net field points out of the page (away from the page).

$A$ very long conducting wire is bent in a semicircular shape from $A$ to $B$ as shown in the figure. The magnetic field at point $P$ for the steady current configuration is given by:

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