Question

A nichrome wire of resistivity $100\mu \Omega \text{cm}$ has length $10\text{m}$ and area of cross-section $0.02{\text{cm}}^2$. If this wire is connected between the terminals of a $10\text{V}$ battery, then the current flow through it is

• A. $4\text{A}$
• B. $2\text{A}$
• C. $1\text{A}$
• D. $\frac{1}{5}\text{A}$

Answer 1

The correct option is B $2\text{A}$

Resistivity of the material of the wire, $\rho =100\mu \Omega \text{cm}=100\times {10}^{-8}\Omega \text{m}$
Length of wire, $l=10\text{m}$
Area of cross section of wire, $A=0.02{\text{cm}}^2=0.02\times {10}^{-4}{\text{m}}^2$
Voltage across battery, $E=10\text{V}$
The current flowing through the wire can be calculated as:
$I=\frac{V}{R}=\frac{VA}{\rho l}(\text{As}R=\rho \frac{l}{A})$
$=\frac{10\times 0.02\times {10}^{-4}}{100\times {10}^{-8}\times 10}$
$=2\text{A}$
Hence, the correct answer is option (b).