A current loop consists of two identical semicircular parts each of radius $R$, one lying in the $x-y$ plane,

Question

A current loop consists of two identical semicircular parts each of radius $R$, one lying in the $x-y$ plane, and the other in the $x-z$ plane. If the current in the loop is $i$, what will be the resultant magnetic field due to the two semicircular parts at their common centre?

Accepted Answer

1. **Magnetic Field of a Semicircle:** The magnetic field at the center of a full circular loop carrying current $i$ is $B = \frac{\mu_0 i}{2R}$. For a semicircular arc (half a loop), the magnitude is half of this: $B_{semi} = \frac{1}{2} \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{4R}$.
2. **Direction:** The direction of the magnetic field follows the right-hand rule .
   - For the semicircle in the $x-y$ plane, the axis is the $z$-axis, so the field $\vec{B}_1$ is along the $z$-direction ($\hat{k}$).
   - For the semicircle in the $x-z$ plane, the axis is the $y$-axis, so the field $\vec{B}_2$ is along the $y$-direction ($\hat{j}$).
3. **Superposition:** The two fields are perpendicular to each other. The resultant magnitude $B_{net}$ is the vector sum:
   $$B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{\left(\frac{\mu_0 i}{4R}\right)^2 + \left(\frac{\mu_0 i}{4R}\right)^2}$$
   $$B_{net} = \frac{\mu_0 i}{4R} \sqrt{2} = \frac{\mu_0 i}{2\sqrt{2} R}$$ .

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