Chapter 26: Q8Q (page 765)

The following table gives the lengths of three copper rods, their diameters, and the potential differences between their ends. Rank the rods according to (a) the magnitude of the electric field within them, (b) the current density within them, and (c) the drift speed of electrons through them, greatest first.

Short Answer

Expert verified

a) The rank of the rods according to the magnitude of the electric field within them, the greatest first is $E_{1} = E_{2} > E_{3}$

b) The rank of the rods according to the current density within them, the greatest first is $J_{1} = J_{2} > J_{3}$

c) The rank of the rods according to the drift speed of electrons through them, the greatest first is ${(V_{d})}_{1} = {(V_{d})}_{2} > {(V_{d})}_{3}$

Step by step solution

The given data

Three copper rods with their lengths, diameters, and the potential difference between their rods as follow,

Understanding the concept of field, density and drift speed

The electric field is defined as the potential difference per unit length. From Ohm’s law, the potential difference is directly proportional to the current through the conductor. The constant of proportionality is called the resistance of that conductor. The current density is the current across the unit area at a given point in the conductor. The drift velocity of the electrons is the average velocity attained by the electrons.

We use the definition of the electric field in terms of potential difference to rank the rods according to the magnitude of the electric field within them. Then, we can use Ohm’s law and resistance in the formula of current density to rank the rods according to the current density. To rank the rods according to the drift speed of the electrons through them, we can use the formula for drift velocity.

Formulae:

The strength of a field due to present potential difference, $E = V / X$ …(i)

Here,E is the electric field,V is the potential difference,X is the distance.

The current equation from Ohm’s law, $l = V / R$ …(ii)

Here,I is the electric current,V is the potential difference,R is the resistance.

The resistance value of a material, $R = p L / A$ …(iii)

Here,Ris the resistance, p is charge density,L is the length of a conductor, the is cross-section area of A the conductor.

The current density of a uniformly flowing current, $J = l / A$ …(iv)

J is the current density,I is the electric current,A the is cross-section area of the conductor.
The drift velocity of the electrons, $V_{d} = \frac{J}{n e}$ …(v)

$V_{d}$ is the drift velocity of the electrons,J is the current density,n is the number of electrons, ande is the charge on the electron.

(a) Calculation of the rank of the sections due to electric field

For rod 1, the value of the electric field using equation (i) can be given as:

$E_{1} = \frac{V}{L}$

For rod 2, the value of the electric field using equation (i) can be given as:

$E_{2} = \frac{2 V}{2 L} = \frac{V}{L}$

For rod 3, the value of the electric field using equation (i) can be given as:

$E_{3} = \frac{2 V}{3 L} = 0.66 \frac{V}{L}$

So, $E_{1} = E_{2} > E_{3}$

Therefore, the rank of the rods according to the magnitude of the electric field within them, greatest first is $E_{1} = E_{2} > E_{3}$

(b) Calculation of the rank due to magnitude of current density

Substituting the values from equations (ii) and (iii) in equation (iv), we can get the current density equation s follows:

$J = \frac{V}{p L}$ …(vi)

For rod 1, the current density using equation (vi) can be given as:

$J_{1} = \frac{V}{p L}$

For rod 2, the current density using equation (vi) can be given as:

$J_{2} = \frac{2 V}{p (2 L)} = \frac{V}{p L}$

For rod 3, the current density using equation (vi) can be given as:

$J_{3} = \frac{2 V}{p (3 L)} = 0.66 \frac{V}{p L}$

So, $J_{2} = J_{1} > J_{3}$

Therefore, the rank of the rods according to the current density within them, the greatest first is, $J_{2} = J_{1} > J_{3}$

(c) Calculation of the rank due to the drift speed

From equation (v), we can get the speed relation to the current density as follows:

$V_{d} α J$

Hence, from calculations of part (b), we get $(V_{d})_{1} = (V_{d})_{2} > (V_{d})_{3}$

Therefore, the rank of the rods according to the drift speed of the electrons through them, the greatest first is $(V_{d})_{1} = (V_{d})_{2} > (V_{d})_{3}$