Q. 114.0( 5 Votes )

# A copper wire of radius 0.1 mm and resistance 1 kΩ is connected across a power supply of 20 V.

(a) How many electrons are transferred per second between the supply and the wire at one end?

(b) Write down the current density in the wire.

Answer :

a) 1.25×10^{17} b) 6.37×10^{5} A/m^{2}

Given,

Radius of the wire, r= 0.1mm=0.1×10^{-3}m

Resistance of the wire, R= 1000Ω

The voltage across the wire, V= 20*V*

Formula used,

The wire is having a resistance, R and hence the current, I through it can be found from the equation,

Where V denotes the potential difference across the resistance.

We know that the current, I through a resistor can be calculated from the charge, q flowing through it per seconds, or

Where t is the time in seconds for which the current is flown.

But the charge, q can be written in terms of number of electrons, n flowing by the relation,

Where e is the charge of 1 electron= 1.6×10^{-19} C

Also, the current density, J through a material with constant area of cross section A, can be represented as

Solution,

a) From the above equations, eqn.1 can be re-written with the help of eqn.2 and eqn.3 as,

Or

For time t= 1s, and by substituting the given values in the above relation, we can calculate the number of electrons passed in 1s through the copper wire with resistance 1000Ω as,

Hence the number of electrons transferred per second between the supply and the wire at one end is 1.25×10^{17}

b)

To find the current density, we have to calculate the area of cross-section, A of the wire with radius r. Hence

By substituting the value for r, we get the area as,

Also, the current through the wire, from eqn.1, is

Now, by substituting the known values in the expression for current density, eqn.4, we get J as

Hence the current density in the wire is around 6.37×10^{5} A/m^{2}

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