A current I enters a circular coil of radius | vedclass

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(A) Let $l_1$ and $l_2$ be the lengths of the two parts of the conductor,$ABC$ and $ADC$ respectively,and $\rho$ be the resistance per unit length of

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(A) Let $l_1$ and $l_2$ be the lengths of the two parts of the conductor,$ABC$ and $ADC$ respectively,and $ho$ be the resistance per unit length of the conductor.The resistance of the part $ABC$ is $R_1 = ho l_1$.The resistance of the part $ADC$ is $R_2 = ho l_2$.Since the two parts are in parallel,the potential difference across $AC$ is the same for both:$V = I_1 R_1 = I_2 R_2$$I_1 (ho l_1) = I_2 (ho l_2)$$I_1 l_1 = I_2 l_2 \quad \dots (i)$The magnetic field at the center $O$ due to a current $I_1$ flowing in arc $ABC$ is:$B_1 = \frac{\mu_0 I_1 	heta_1}{4 \pi R} = \frac{\mu_0 I_1 l_1}{4 \pi R^2}$ (directed into the page,$\otimes$)The magnetic field at the center $O$ due to a current $I_2$ flowing in arc $ADC$ is:$B_2 = \frac{\mu_0 I_2 	heta_2}{4 \pi R} = \frac{\mu_0 I_2 l_2}{4 \pi R^2}$ (directed out of the page,$\odot$)Using equation $(i)$,$I_1 l_1 = I_2 l_2$,we get:$B_2 = \frac{\mu_0 I_1 l_1}{4 \pi R^2}$Since $B_1$ and $B_2$ are equal in magnitude and opposite in direction,the resultant magnetic field at the center $O$ is:$B = B_2 - B_1 = 0$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{r} \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$
$(B). \frac{\mu_0 i}{2r} \odot$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$
$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$
$(D). \frac{\mu_0 i}{4r} \odot$	$(D). 0$	$(D). 0$

$A$ current $I$ enters a circular coil of radius $R$,branches into two parts and then recombines as shown in the circuit diagram. The resultant magnetic field at the centre of the coil is

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