If a battery of 12 text{ V} is connected acro | vedclass

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(A) When the battery is connected across the diametrically opposite points $A$ and $B$,the ring is divided into two equal semicircular arcs,$ACB$ and

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(A) When the battery is connected across the diametrically opposite points $A$ and $B$,the ring is divided into two equal semicircular arcs,$ACB$ and $ADB$.Since the arcs are in parallel,the potential difference across both is the same $(12 	ext{ V})$.Let $R_1$ and $R_2$ be the resistances of the two semicircular parts. Since they have equal lengths and cross-sectional areas,$R_1 = R_2 = R_{arc}$.By Ohm's law,the current $I_1$ through arc $ACB$ and $I_2$ through arc $ADB$ will be equal because the resistances are equal $(I_1 = I_2 = I/2)$.The magnetic field at the centre due to a semicircular arc carrying current $I'$ is $B = \frac{\mu_0 I'}{4R}$.The magnetic field due to arc $ACB$ $(B_1)$ at the centre is $\frac{\mu_0 (I/2)}{4R}$ directed into the plane (using the right-hand rule).The magnetic field due to arc $ADB$ $(B_2)$ at the centre is $\frac{\mu_0 (I/2)}{4R}$ directed out of the plane.Since the magnitudes are equal and the directions are opposite,the net magnetic field at the centre is $B_{net} = B_1 - B_2 = 0$.

If a battery of $12 \text{ V}$ is connected across the diametrically opposite points $A$ and $B$ of a conducting ring of radius $R$ and the current drawn from the battery is $I$,then the magnetic field produced at the centre of the ring due to the ring is . . . . . . .

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