Q.1. Two wires each of length 40 m and diameter 2.6 mm, are connected in series. A potential difference of 50 V is applied between the ends of the composite wire. The resistances the wires are 0.13 Ω and 0.75 Ω, respectively. Determine:
i) the current density in each wire;
ii) the potential difference across each wire;
iii) the magnitude of the electric field in each wire;
iv) the resistivities of the two wires;
v) identify the wire materials.
i) To determine the current density in each wire, we need to calculate the current flowing through each wire and divide it by the cross-sectional area of the wire.
The formula for current density (J) is given by:
J = I / A
Where J is the current density, I is the current, and A is the cross-sectional area.
First, let's calculate the current flowing through the composite wire. Since the wires are connected in series, the current flowing through both wires will be the same.
Given:
Length of each wire (L) = 40 m
Diameter of each wire (d) = 2.6 mm = 0.0026 m
Potential difference (V) = 50 V
Resistance of the first wire (R1) = 0.13 Ω
Resistance of the second wire (R2) = 0.75 Ω
To find the total resistance (R) of the composite wire, we add the resistances of the individual wires:
R = R1 + R2
Let's calculate the total resistance:
R = 0.13 Ω + 0.75 Ω
R = 0.88 Ω
Using Ohm's Law, we can calculate the current (I):
I = V / R
I = 50 V / 0.88 Ω
I ≈ 56.82 A
Now, let's calculate the cross-sectional area (A) of each wire using the formula:
A = π * (d/2)^2
For the first wire:
A1 = π * (0.0026 m / 2)^2
For the second wire:
A2 = π * (0.0026 m / 2)^2
Now, we can calculate the current density for each wire using the formula mentioned earlier:
J1 = I / A1
J2 = I / A2
ii) Now, let's calculate the potential difference across each wire.
Since the wires are in series, the total potential difference across the composite wire is equal to the sum of the potential differences across each wire.
V1 = I * R1
V2 = I * R2
iii) To find the magnitude of the electric field in each wire, we can use Ohm's Law:
E = V / L
Where E is the electric field, V is the potential difference, and L is the length of the wire.
E1 = V1 / L
E2 = V2 / L
iv) The resistivity (ρ) of a material can be calculated using the formula:
ρ = (R * A) / L
Where ρ is the resistivity, R is the resistance, A is the cross-sectional area, and L is the length of the wire. Using this formula, we can find the resistivity of each wire.
For the first wire:
ρ1 = (R1 * A1) / L
For the second wire:
ρ2 = (R2 * A2) / L
v) To identify the wire materials, we need to compare the resistivities of the wires with the resistivity values of known materials. Each material has its own unique resistivity value.
To answer these questions, we need to use some formulas and equations related to electrical resistance, current, and resistivity. Let's break it down step by step:
Step 1: Calculate the total resistance of the composite wire.
The resistances of the two wires are given as 0.13 Ω and 0.75 Ω. Since they are connected in series, the total resistance (R_total) can be calculated by adding them together:
R_total = 0.13 Ω + 0.75 Ω = 0.88 Ω
Step 2: Calculate the current flowing through the composite wire.
We are given a potential difference (V) of 50 V applied between the ends of the composite wire. According to Ohm's law (V = IR), we can find the current (I) using the obtained total resistance (R_total):
I = V / R_total = 50 V / 0.88 Ω = 56.82 A
Step 3: Calculate the cross-sectional area of each wire.
The diameter of each wire is given as 2.6 mm. We can calculate the radius (r) using the formula: r = diameter / 2.
For each wire, r = 2.6 mm / 2 = 1.3 mm = 0.0013 m.
The cross-sectional area (A) of each wire can be found using the formula: A = πr^2.
For each wire, A = π(0.0013 m)^2 ≈ 5.309291547 m^2.
Step 4: Calculate the current density in each wire.
Current density (J) is defined as the current passing through a unit area of a conductor. It can be calculated using the formula: J = I / A.
For each wire:
J = 56.82 A / 5.309291547 m^2 ≈ 10.7 A/m^2.
Step 5: Calculate the potential difference across each wire.
Since the wires are connected in series, the potential difference across each wire will be the same as the total applied potential difference, which is 50 V.
Step 6: Calculate the magnitude of the electric field in each wire.
The magnitude of the electric field (E) in a wire is given by Ohm's law (E = V / L), where V is the potential difference and L is the length of the wire. Since the potential difference across each wire is 50 V, we need to calculate the lengths of each wire. It is given that both wires have a length of 40 m, so E will be the same for both wires:
E = 50 V / 40 m = 1.25 V/m.
Step 7: Calculate the resistivity of each wire.
The resistivity (ρ) of a material can be calculated using the formula:
ρ = RA / L,
where R is the resistance of the wire, A is the cross-sectional area, and L is the length of the wire.
For the first wire:
ρ1 = (0.13 Ω)(5.309291547 m^2) / 40 m ≈ 0.017 Ω·m.
Similarly, for the second wire:
ρ2 = (0.75 Ω)(5.309291547 m^2) / 40 m ≈ 0.099 Ω·m.
Step 8: Identify the wire materials.
To identify the wire materials, we need to compare the obtained values of resistivity (ρ) with the known values for different materials. Comparing ρ1 ≈ 0.017 Ω·m with known values, it suggests that the first wire is made of Copper. Similarly, comparing ρ2 ≈ 0.099 Ω·m, it suggests that the second wire is made of Aluminum.
So, to summarize the answers:
i) The current density in each wire is approximately 10.7 A/m^2.
ii) The potential difference across each wire is 50 V.
iii) The magnitude of the electric field in each wire is approximately 1.25 V/m.
iv) The resistivities of the two wires are approximately 0.017 Ω·m for Copper and 0.099 Ω·m for Aluminum.
v) The wire materials are Copper and Aluminum.