Question

A cylindrical copper cable 1.90km long is connected across a 220.0V -potential difference. 

What should be its diameter so that it produces heat at a rate of 70.0W ? 

What is the electric field inside the cable under these conditions? 

Answer 1

Concepts and reason

The concepts required to solve the given question are power rule, Ohm’s law, relation between the resistance and diameter of the wire and the relation between the electric field and electric potential.

First, find the relation between the power, voltage and resistance. Using the above relation calculate the resistance of the wire. From the expression for the resistance of the wire, radius of the wire can be calculated. Multiply it with two gives the diameter of the wire. To find the electric field inside the wire, use the relation between the electric field, electric potential and the length of the wire. Thus, the electric field can be calculated.

Fundamentals

The Ohm’s law gives the relation between current, voltage and resistance. The law states that the “electric current is directly proportional to the voltage and inversely proportional to the resistance of the circuit”.

Expression for Ohm’s law is given by,

Here, is the voltage, is the current, and is the resistance of the circuit.

Electric current is the flow of electric charge. Electrical resistance is the resistance to the flow of electric current. Voltage is defined as the electrical potential difference between two points.

Resistance of wire is given by,

$R= $

Here, is the resistance, is the resistivity, is the length of the wire, and is the area of the wire.

Power is defined as the “rate of doing work per unit time”. The mathematical expression for power rule is given by,

$P=IV $

Here, is the power dissipated, is the current and is the voltage.

Electric field is defined as the “negative gradient of scalar potential”. Scalar potential is the “Amount of work done in moving a small charge from infinity to a particular point”. Potential difference is the change in potential between two points.

Expression for electric field between the plates is,

Here, is the electric field between the plates, is the potential difference, and is the distance.

Power rule is given by,

$P=IV $

Substitute for I in the above expression.

Resistance of the wire is,

$R= $

Substitute for and for .

Consider the expression for resistance of the wire,

$R= $

The area of the wire is,

Substitute for , for , and for .

Consider the expression for the area of the wire,

Radius of the wire,

Substitute for and for .

Diameter of the wire is,

$D=2r $

Substitute for .

Electric field inside the cable is,

Here, the distance can be taken as the length of the wire. Hence the above expression can be written as,

Substitute for and for .

Ans:

The diameter of the copper wire is.

The electric field inside the cable is .