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Question

(A) Let the total current be $i$. The wire is divided into two parts $ABC$ and $ADC$ connected in parallel.The resistance of a wire is proportional to

Vedclass · Accepted Answer

$\frac{\mu_0}{2R} \frac{i l_1 l_2}{(l_1+l_2)^2}$

Vedclass · Answer

(A) Let the total current be $i$. The wire is divided into two parts $ABC$ and $ADC$ connected in parallel.The resistance of a wire is proportional to its length $(R \propto l)$. Let $r_1$ and $r_2$ be the resistances of parts $ABC$ and $ADC$ respectively.$r_1 = ho \frac{l_1}{A}$ and $r_2 = ho \frac{l_2}{A}$, where $ho$ is resistivity and $A$ is cross-sectional area.The current $i_1$ in part $ABC$ is given by the current divider rule:$i_1 = i \left( \frac{r_2}{r_1 + r_2} ight) = i \left( \frac{l_2}{l_1 + l_2} ight)$.The magnetic field at the center due to an arc of length $l_1$ carrying current $i_1$ is:$B = \frac{\mu_0 i_1 	heta}{4\pi R}$, where $	heta$ is the angle subtended by the arc at the center.Since the circumference is $L = l_1 + l_2 = 2\pi R$, the angle $	heta = \frac{l_1}{R}$.Substituting $i_1$ and $	heta$:$B = \frac{\mu_0}{4\pi R} \left( i \frac{l_2}{l_1 + l_2} ight) \left( \frac{l_1}{R} ight) = \frac{\mu_0 i l_1 l_2}{4\pi R^2 (l_1 + l_2)}$.Wait, checking the options provided, the standard result for this specific problem type often simplifies to the form in option $(A)$ if we consider the magnetic field contribution of the arc as $B = \frac{\mu_0 i_1}{2R} \frac{	heta}{2\pi}$.Using $B = \frac{\mu_0 i_1}{2R} \frac{l_1}{2\pi R} = \frac{\mu_0}{2R} \left( i \frac{l_2}{l_1+l_2} ight) \frac{l_1}{l_1+l_2} = \frac{\mu_0 i l_1 l_2}{2R(l_1+l_2)^2}$.

$A$ uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the centre due to $ABC$ portion of the wire will be (length of $ABC = l_1$, length of $ADC = l_2$):

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