A uniform wire is bent in the form of a circle of radius $R$. A current $I$ enters at $A$ and leaves at $C$ as shown in the figure :If the length $ABC$ is half of the length $ADC,$ the magnetic field at the centre $O$ will be
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(a) Directions of currents in two parts are different, so directions of magnetic fields due to these currents are opposite. Also applying Ohm’s law across $AB$
${i_1}{R_1} = {i_2}{R_2} \Rightarrow {i_1}{l_2} = {i_2}{l_2}$ $\left( {\;R = \rho \frac{l}{A}} \right)$
Also ${B_1} = \frac{{{\mu _o}}}{{4\pi }} \times \frac{{{i_1}{l_1}}}{{{r^2}}}$ and ${B_2} = \frac{{{\mu _o}}}{{4\pi }} \times \frac{{{i_2}{l_2}}}{{{r^2}}}$ ($\;l = r\theta $)
$\therefore \,\frac{{{B_2}}}{{{B_1}}} = \frac{{{i_1}{l_1}}}{{{i_2}{l_2}}} = 1$
Hence, two field induction’s are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of $\theta $.
${i_1}{R_1} = {i_2}{R_2} \Rightarrow {i_1}{l_2} = {i_2}{l_2}$ $\left( {\;R = \rho \frac{l}{A}} \right)$
Also ${B_1} = \frac{{{\mu _o}}}{{4\pi }} \times \frac{{{i_1}{l_1}}}{{{r^2}}}$ and ${B_2} = \frac{{{\mu _o}}}{{4\pi }} \times \frac{{{i_2}{l_2}}}{{{r^2}}}$ ($\;l = r\theta $)
$\therefore \,\frac{{{B_2}}}{{{B_1}}} = \frac{{{i_1}{l_1}}}{{{i_2}{l_2}}} = 1$
Hence, two field induction’s are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of $\theta $.
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