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(A) The magnetic field at the center of a circular arc of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i 	heta}{4\pi r}$.Since the tw

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Zero

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(A) The magnetic field at the center of a circular arc of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i 	heta}{4\pi r}$.Since the two parts of the circular conductor are connected in parallel to the cell,the potential difference $V$ across them is the same.Let $R_1$ and $R_2$ be the resistances of the two parts with lengths $l_1$ and $l_2$ respectively. Then $V = i_1 R_1 = i_2 R_2$.Since $R = ho \frac{l}{A}$,we have $i_1 ho \frac{l_1}{A} = i_2 ho \frac{l_2}{A}$,which implies $i_1 l_1 = i_2 l_2$.The magnetic field due to the first part is $B_1 = \frac{\mu_0 i_1 	heta_1}{4\pi r}$ and due to the second part is $B_2 = \frac{\mu_0 i_2 	heta_2}{4\pi r}$.Since $l_1 = r 	heta_1$ and $l_2 = r 	heta_2$,we have $i_1 r 	heta_1 = i_2 r 	heta_2$,so $i_1 	heta_1 = i_2 	heta_2$.Thus,$B_1 = \frac{\mu_0}{4\pi r} (i_1 	heta_1)$ and $B_2 = \frac{\mu_0}{4\pi r} (i_2 	heta_2)$.Since $i_1 	heta_1 = i_2 	heta_2$,the magnitudes are equal $(B_1 = B_2)$.Because the currents flow in opposite directions around the center,the magnetic fields produced by the two parts are equal in magnitude and opposite in direction.Therefore,the resultant magnetic field at the center is zero.

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

$A$ cell is connected between two points of a uniformly thick circular conductor. The magnetic field at the centre of the loop will be

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