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 $\frac{μ_{0} I}{R}$ times ( MA = R, MB = 2R, ∠DMA = 90^o )

$\frac{5}{16}$ , 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.

Solution

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

Magnetic field at the centre M due to current through the curved portion DA is

$\vec{B_{1}} = \frac{μ_{0} I}{4 πR} \times (\frac{3 π}{2})$

= $\frac{μ_{0} I}{16 R}$

Magnetic field at the centre M due to current through the straight portion CD is B,= 0, since
point M lies on the axis of the straight portion CD.

The resultant magnetic field at the point M is

$\vec{B} = \vec{B_{1}} + \vec{B_{2}} + \vec{B_{3}} + \vec{B_{4}}$

= $(\frac{3 μ_{0} I}{8 R} + 0 + \frac{μ_{0} I}{16 R} + 0)$

⇒ $\vec{B} = \frac{7 μ_{0} I}{16 R}$

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