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Physics-

General

Easy

Question

A current $i$ is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.
The magnetic field at the centre of the loop is $fraction numerator mu subscript 0 end subscript i over denominator R end fraction$ times.
$left parenthesis M A equals R comma blank M B equals 2 R comma blank angle D M A equals 90 degree$

$\frac{5}{16},$ but out of the plane of the paper
$\frac{5}{16},$ but into the plane of the paper
$\frac{7}{16},$ but out of the plane of the paper
$\frac{7}{16},$ but into the plane of the paper

The correct answer is: $fraction numerator 7 over denominator 16 end fraction comma$ but into the plane of the paper

(i) Magnetic field at the centre due to the curved portion $D A equals fraction numerator mu subscript 0 end subscript i over denominator 4 pi R end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses$
According to right hand screw rule, the magnetic field will be into the plane of paper.

(ii) Magnetic field at $M$ due to $A B$ is zero.
(iii) Magnetic field at the centre due to the curved portion $B C$ is $fraction numerator mu subscript 0 end subscript i over denominator 4 pi 2 R end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses.$ According to
right hand screw rule, the magnetic field will be into the plane of paper.
(iv) Magnetic field at $M$ due to $D C$ is zero.
Hence, the resultant magnetic field at $M$
$equals fraction numerator 3 mu subscript 0 end subscript i over denominator 8 R end fraction plus 0 plus fraction numerator mu subscript 0 end subscript i over denominator 16 R end fraction plus 0 equals fraction numerator 7 mu subscript 0 end subscript i over denominator 16 R end fraction$

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