Question

The current in the primary circuit of a potentiometer is 0.2A. The specific resistance and cross-section of the potentiometer wire are 4×10−7ohm meter and 8×10−7m2 respectively. The potential gradient will be equal to:

Answer 1

Final Answer : 0.1 V/m

Steps:
1) Potential Gradient,x : Potential drop per unit length.
$x= \frac{v}{l}$

Where V is Potential Difference
L. : Length of wire
A : Area of Cross section of wire.
I, Current = 0.2A

2) We know that,
Resistance = pl/A
p -> specific resistance
l -> length of wire.
A -> Area of cross section.

Use it in equation (1),
$x = \frac{ir}{l} = \frac{ipl}{la} = \frac{ip}{l} \\ = > x = \frac{0.2 \times 4 \times {10}^{ - 7} }{8 \times {10}^{ - 7} } \\ = > x = 0.1 \: units$

Therefore, Potential Gradient is 0.1 V/m.

Answer 2

Answer:

Potential gradient=v/L

Phi =P *i / A

Given that P=4×10^-7. i=0.2A

A=8×10^-7

Phi =P*i/A

=( 4*10^-7)×0.2÷ (8*10^-7)

= 0 .8 / 8

Phi = 0.1 v/ m