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(B) Let $R_{AB} = 4\,\Omega$ be the resistance of the wire $AB$. Let $R_{AJ}$ be the resistance of the segment $AJ$. The potential difference across $

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$0.3$

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(B) Let $R_{AB} = 4\,\Omega$ be the resistance of the wire $AB$. Let $R_{AJ}$ be the resistance of the segment $AJ$. The potential difference across $AJ$ is given by $V_{AJ} = I \cdot R_{AJ}$,where $I$ is the current in the main circuit.For the first case,the current in the main circuit is $I_1 = \frac{6}{R_{h1} + R_{AB}} = \frac{6}{2 + 4} = 1\,	ext{A}$.The potential drop across $AJ$ is $V_{AJ} = I_1 \cdot R_{AJ} = 1 \cdot R_{AJ} = \varepsilon_1 = 0.5\,	ext{V}$.Thus,$R_{AJ} = 0.5\,\Omega$.For the second case,the current in the main circuit is $I_2 = \frac{6}{R_{h2} + R_{AB}} = \frac{6}{6 + 4} = 0.6\,	ext{A}$.Since the null point $J$ is the same,the resistance $R_{AJ}$ remains $0.5\,\Omega$.The emf $\varepsilon_2$ is equal to the potential drop across $AJ$ in the second case:$\varepsilon_2 = I_2 \cdot R_{AJ} = 0.6 	imes 0.5 = 0.3\,	ext{V}$.

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