Question

A Nichrome wire 30 cm long and 0.25 mm in diameter is connected to a 1.5 V flashlight battery. What is the electric field inside the wire? Why you don’t have to know how the wire is bent? How would your answer change if the wire diameter change were 0.35 mm? (Not that the electric field in the wire is quiet small compared to the electric field near a charged tape.)

Short answer

The electric field inside the wire is $5\text{V}/\text{m}$ and it remains same for new wire diameter.

Step-by-step solution

Identification of given data

The emf of the flashlight battery is $\epsilon =1.5\text{V}$

The length of Nichrome wire is $l=30\text{cm}$

The diameter of wire is $d=0.25\text{mm}$

The new diameter of wire is $D=0.35\text{mm}$
The amount of charge at the positive end of battery plate is found by the product of the capacitance and emf of the battery.

Determination of electric field inside the wire

The electric field inside the wire is given as:

$E=\frac{V}{l}$

Substitute all the values in above equation.

$$\begin{gathered} E=\frac{1.5\text{V}}{\left(30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)} \\ E=5\text{V}/\text{m} \end{gathered}$$

The electric field does not change with the diameter of the wire because the length of wire and emf of flashlight battery is constant.

Therefore, the electric field inside the wire is $5\text{V}/\text{m}$ and it remains same for new wire diameter.