Correct Answer: Option A
Hint: Resultant magnetic field is equal to sum of the magnetic field due to the currents in the two semicircles. This can be found using the equation of magnetic field at the centre of a loop.
Explanation of Correct Option:
Step 1: Finding magnetic field due to the two loops.
Let the magnetic field due to the semicircular loop of radius $$R_{1}$$ be $$B_{1}$$ and magnetic field due to the semicircular loop of radius $$R_{2}$$ be $$B_{2}$$
Resultant magnetic field at $$O=$$ Magnetic field due to the semicircular loop of radius $$R_{1}$$ + Magnetic field due to the semicircular loop of radius $$R_{2}$$
$$=B_{1}+B_{2}$$
Magnetic field at the centre of a current carrying loop $$= \dfrac{\mu_{0}i}{2\ r}$$
Magnetic field at the centre of a semicircular current carrying loop $$=\dfrac{1}{2}\times \dfrac{\mu_{0}i}{2\ r} = \dfrac{\mu_{0}i}{4\ r}$$
$$B_{1}=\dfrac{\mu_{0}i}{4\ R_{1}}$$
$$B_{2}=-\dfrac{\mu_{0}i}{4\ R_{2}}$$ (Negative sign since the direction is opposite)
Step 2: FInding the magnetic field due to horizontal parts
We know,
$$dB=\dfrac{\mu i}{4\pi}(\dfrac{\vec{dl}\hat{r}sin\theta} {r^{2}})$$
Here since the angle is $$0^{\circ}$$ and $$180^{\circ}$$, magnetic field due to the horizontal parts is zero.
Step 3: Finding magnetic induction at the centre.
Magnetic induction at the centre $$O, B= B_{1}+B_{2}$$
$$B=\dfrac{\mu_{0}i}{4\ R_{1}}+-\dfrac{\mu_{0}i}{4\ R_{2}}$$
$$B=\dfrac{\mu_{0}i}{4}(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$$
The magnetic induction at centre in the shown figure is $$B=\dfrac{\mu_{0}i}{4}(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}).$$