Given,
Length of potentiometer wire $(l) = 10\;m$
Resistance of potentiometer $(R_p) = 20\; Ω$
Potential of battery $(V_b) = 3V$
Resistance of battery $(R_b) = 10 \; Ω$
Potential Gradient $(\frac{V}{l}) = \;?$
Let $I$ be the current flowing through the circuit.
Then, from Ohm's law,
$V_b = IR$ [where R= Total resistance of the circuit = $R_b + R_p$
or, $3 = I\;*\;(10+20)$
⇒ I = 0.1 A
Since the combination is in series, the current (I) all over the circuit is same.
So, p.d in AB $(V_{AB})$ = I * $R_{b}$ = 0.1 * 20 = 2 V
Now, Potential Gradient $(\frac{V}{l}) = \frac{2}{10}$ = 0.2V/m
Hence, the potential gradient along the wire is 0.2 V/m.
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