Question

The current \(I\) is flowing through a loop \(A B C D A\) as shown in the figure. The magnitude of the magnetic field at the centre ' \(O\) ' is \(x\) times

\[ \left(\frac{\mu_0 \mathrm{I}}{\mathrm{R}}\right)\left(\mathrm{OD}=\mathrm{R}, \mathrm{DC}=\mathrm{R}, \angle \mathrm{AOD}=90^{\circ}\right) \]

The value of ' \(x\) ' is ( \(\mu_0=\) permeability of free space)

• A. \(\frac{5}{16}\)
• B. \(\frac{5}{12}\)
• C. \(\frac{7}{16}\)
• D. \(\frac{7}{12}\)

Hint:

Straight segments along a radius through O give no field at O; only the arcs contribute, each as a fraction of a full circle field.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Field at the centre of a circular arc is \(B = \dfrac{\mu_0 I}{2r}\times\dfrac{\theta}{2\pi}\). A straight wire whose line passes through O gives zero field at O.

Step 2: Straight Parts
Segments AB and CD both lie along lines through O, so they contribute nothing.

Step 3: Arcs
Arc DA has radius \(R\) and subtends \(270^{\circ}\): \(B_1 = \dfrac{3}{4}\cdot\dfrac{\mu_0 I}{2R} = \dfrac{3\mu_0 I}{8R}\).
Arc BC has radius \(2R\) and subtends \(90^{\circ}\): \(B_2 = \dfrac14\cdot\dfrac{\mu_0 I}{2(2R)} = \dfrac{\mu_0 I}{16R}\).

Step 4: Total
Both arcs carry current anticlockwise, so the fields add:
\(B = \dfrac{\mu_0 I}{R}\left(\dfrac38 + \dfrac1{16}\right) = \dfrac{7}{16}\dfrac{\mu_0 I}{R}\), so \(x = \dfrac{7}{16}\), option C.

Final Answer:
The value of x is 7 by 16. \[ \boxed{\text{(C) }\dfrac{7}{16}} \]