question_answer

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 \[\frac{{{\mu }_{0}}i}{R}\] times  (\[MA=R\], \[MB=2R\],\[\angle DMA=90{}^\circ \])

A)  \[\frac{5}{16},\]but out of the plane of the paper

B)  \[\frac{5}{16},\]but into the plane of the paper

C)  \[\frac{7}{16},\]but out of the plane of the paper

D)  \[\frac{7}{16},\]but into the plane of the paper

Correct Answer: D

Solution :

(i) Magnetic field at the centre due to the curved portion \[DA=\frac{{{\mu }_{0}}i}{4\pi R}\left( \frac{3\pi }{2} \right)\] According to right hand screw rule, the magnetic field will be into the plane of paper. (ii) Magnetic field at M due to AB is zero. (iii) Magnetic field at the centre due to the curved  portion  BC  is  \[\frac{{{\mu }_{0}}i}{4\pi 2R}\left( \frac{\pi }{2} \right).\] According to right hand screw rule, the magnetic field will be into the plane of paper. (iv) Magnetic field at M due to DC is zero. Hence, the resultant magnetic field at M \[=\frac{3{{\mu }_{0}}i}{8R}+0+\frac{{{\mu }_{0}}i}{16R}+0=\frac{7{{\mu }_{0}}i}{16R}\]