A current $I$ flowing through the loop as shown in the adjoining figure. The magnetic field at center O is

\frac{17 μ_{0} I}{16 R} \otimes

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\frac{17 μ_{0} I}{16 R} ⊙

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\frac{5 μ_{0} I}{16 R} \otimes

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\frac{5 μ_{0} I}{16 R} ⊙

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Solution

The correct option is B $\frac{17 μ_{0} I}{16 R} ⊙$
No correct option.

Given,

A loop in which current is flowing,

So, Magnetic field due to part 1 is: $B_{1} = \frac{μ_{0} I}{2 R} \times \frac{3}{4}$

The magnetic field due to part 2 is: $B_{2} = \frac{μ I}{2 (2 R)} \times \frac{1}{4}$

So the total magnetic field is: $B = B_{1} + B_{2} \Rightarrow \frac{μ_{0}}{R} (\frac{3}{8} + \frac{1}{16}) = \frac{7 μ_{0} I}{16 R}$

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