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Question: A wire has a diameter of \(0.2\,mm\) and a length of \(50\,cm\). The specific resistance of its mate...

A wire has a diameter of 0.2 mm0.2\,mm and a length of 50 cm50\,cm. The specific resistance of its material is 40×10−6 ohm cm40 \times {10^{ - 6}}\,ohm\,cm. The current through it, when a potential difference of 2 V2\,V is applied across it is
A. 3.14 A3.14\,A
B. 31.4 A31.4\,A
C. 0.314 A0.314\,A
D. 0.0314 A0.0314\,A

Explanation

Solution

Hint- We can use ohm’s law to find the current flowing in the wire. According to ohm’s law voltage is directly proportional to current.
The resistance of the wire is directly proportional to the length of the wire and inversely proportional to the area of the cross section of the wire.
Using this information, we can find the value of current through the wire.

Complete step by step answer:
It is given that the diameter of a wire,
d=0⋅2 mmd = 0 \cdot 2\,mm
⇒d=0⋅2×10−3m\Rightarrow d = 0 \cdot 2 \times {10^{ - 3}}m
Radius is half the diameter.
⇒r=0⋅1×10−3m\Rightarrow r = 0 \cdot 1 \times {10^{ - 3}}m
The length of the wire is,
l=50 cml = 50\,cm
⇒l=50×10−2m\Rightarrow l = 50 \times {10^{ - 2}}m
The specific resistance of its material is
ρ=40×10−6 ohm cm\rho = 40 \times {10^{ - 6}}\,ohm\,cm
We need to calculate the current through the wire.
The potential difference applied across the wire is given as,
V=2 VV = 2\,V
In order to find the current, we can use ohm’s law.
According to ohm’s law the voltage in a circuit is directly proportional to the current flowing through
the circuit.
That is,
V∝IV \propto I
⇒V=IR\Rightarrow V = IR
Where, I is the current, R is the resistance.
The resistance is the obstruction to the flow of current. It is directly proportional to the length and inversely proportional to the area of cross section.
R∝lAR \propto \dfrac{l}{A}
⇒R=ρlA\Rightarrow R = \dfrac{{\rho l}}{A}
Where, ρ\rho is a constant called the resistivity of the material.
On substituting equation (2) in equation (1), we get
v=IρlAv = \dfrac{{I\rho l}}{A}
From this we can get the equation for calculating current as,
⇒I=VAρl\Rightarrow I = \dfrac{{VA}}{{\rho l}}
The cross-sectional area of wire is the area of a circle
⇒A=πr2\Rightarrow A = \pi {r^2}.
Now let us substitute all the given values
⇒I=2×π×(0⋅1×10−3)240×10−8×50×10−2\Rightarrow I = \dfrac{{2 \times \pi \times {{\left( {0 \cdot 1 \times {{10}^{ - 3}}} \right)}^2}}}{{40 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 2}}}}
∴I=0⋅314A\therefore I = 0 \cdot 314A
This is the value of current flowing in the wire.

So, the correct answer is option C.

Note: - Ohm’s law which static voltage is directly proportional to current is valid only when the temperature and other physical conditions are kept constant. This law is not applicable when elements such as diode, transistor etc are used. This is because resistance of elements like these changes with temperature.