A straight conductor carrying current $I$ splits into two parts as shown in the figure. The radius of the circ

Question

A straight conductor carrying current $I$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is:

Accepted Answer

1. **Concept:** The problem involves determining the net magnetic field at the center of a circular loop where the current splits into two parallel paths (arcs). The two arcs have lengths $l_1$ and $l_2$ and carry currents $I_1$ and $I_2$ respectively.
2. **Current Division:** Since the two arcs are connected in parallel, the potential difference ($V$) across them is the same. According to Ohm's Law, $V = I_1 R_1 = I_2 R_2$. Since resistance $R$ is proportional to length $l$ for a uniform wire ($R = \rho l/A$), we have $I_1 l_1 = I_2 l_2$ .
3. **Magnetic Field Calculation:** The magnitude of the magnetic field $B$ at the center of a circular arc of length $l$ radius $R$ is given by the Biot-Savart Law as $B = \frac{\mu_0 I l}{4\pi R^2}$ .
   - Field due to arc 1: $B_1 = \frac{\mu_0 I_1 l_1}{4\pi R^2}$
   - Field due to arc 2: $B_2 = \frac{\mu_0 I_2 l_2}{4\pi R^2}$
4. **Superposition:** The currents in the two arcs flow in opposite senses (one clockwise, the other counter-clockwise). Therefore, their magnetic fields at the center are directed in opposite directions (one inward, one outward). Since $I_1 l_1 = I_2 l_2$, the magnitudes $B_1$ and $B_2$ are equal.
   $$B_{net} = B_1 - B_2 = 0$$

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