Figure shows two semicircular loops of radii $R_1$ and $R_2$ carrying current $I$. The magnetic field at the common centre '$O$' is
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The magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2R}$. For a semicircle, it is exactly half of that, i.e., $B = \frac{\mu_0 I}{4R}$.
Step 1: The magnetic field $B$ produced by a circular arc carrying current $I$ at its center of curvature is given by:
\[ B = \frac{\mu_0 I \theta}{4\pi R} \]
where $\theta$ is the angle subtended by the arc in radians.
Step 2: For a semicircular arc, $\theta = \pi$ radians. Thus, the field at the center is:
\[ B = \frac{\mu_0 I \pi}{4\pi R} = \frac{\mu_0 I}{4R} \]
Step 3: According to the figure, both semicircular loops of radii $R_1$ and $R_2$ carry current $I$ in a way that their magnetic fields at the common center $O$ are in the same direction (into the page or out of it based on the right-hand thumb rule).
Step 4: The total magnetic field $B_{eq}$ is the sum of the fields produced by each semicircle:
\[ B_{eq} = B_1 + B_2 \]
\[ B_{eq} = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2} \]
Step 5: Factoring out the common terms:
\[ B_{eq} = \frac{\mu_0 I}{4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]
