Consider a potentiometer circuit where the primary loop contains a...

Let the total length of the potentiometer wire be $L = 600 \text{ cm}$ and its total resistance be $R = 15r$.
The jockey divides the wire into two segments at a distance $l = 560 \text{ cm}$ from point A. Let the point where the jockey touches be J, and the other end of the wire be B.
The resistance of the segment AJ is:
$R_{AJ} = \dfrac{l}{L} \times R = \dfrac{560}{600} \times 15r = 14r$
The resistance of the remaining segment JB is:
$R_{JB} = 15r - 14r = r$

We can simplify the primary circuit using Thevenin's theorem across the points A and J. First, we find the open-circuit voltage ($V_{th}$) and Thevenin resistance ($R_{th}$) by removing the secondary branch.
The primary circuit is a single loop containing the battery $\mathcal{E}$, its internal resistance $r$, and the potentiometer wire of resistance $15r$.
The total resistance of this primary loop is $R_{primary} = r + 15r = 16r$.
The current in the primary loop is:
$I_0 = \dfrac{\mathcal{E}}{16r}$
The open-circuit potential difference across AJ is:
$V_{th} = V_A - V_J = I_0 \times R_{AJ} = \left(\dfrac{\mathcal{E}}{16r}\right) \times 14r = \dfrac{7\mathcal{E}}{8}$

To find the Thevenin resistance $R_{th}$ across A and J, we short-circuit the battery $\mathcal{E}$. The resistance $R_{th}$ is the parallel combination of $R_{AJ}$ and the rest of the primary loop ($R_{JB} + r$):
$R_{th} = \dfrac{R_{AJ} \times (R_{JB} + r)}{R_{AJ} + (R_{JB} + r)} = \dfrac{14r \times (r + r)}{14r + 2r} = \dfrac{28r^2}{16r} = \dfrac{7r}{4}$

Now, we connect the secondary branch across the Thevenin equivalent circuit. The secondary branch contains a battery of emf $\mathcal{E}/2$ and a resistor $r$. Since the positive terminals of both batteries are connected at A, the batteries oppose each other in the equivalent loop.
The net EMF in the loop is:
$\mathcal{E}_{net} = V_{th} - \dfrac{\mathcal{E}}{2} = \dfrac{7\mathcal{E}}{8} - \dfrac{\mathcal{E}}{2} = \dfrac{3\mathcal{E}}{8}$
The total resistance in the equivalent loop is:
$R_{total} = R_{th} + r = \dfrac{7r}{4} + r = \dfrac{11r}{4}$

The current $I$ through the galvanometer is:
$I = \dfrac{\mathcal{E}_{net}}{R_{total}} = \dfrac{\dfrac{3\mathcal{E}}{8}}{\dfrac{11r}{4}} = \dfrac{3\mathcal{E}}{8} \times \dfrac{4}{11r} = \dfrac{3\mathcal{E}}{22r}$

Answer: $\dfrac{3\mathcal{E}}{22 r}$