Answer:
16.5 kwh and 59400 kJ.
Explanation:
kWh is a measure of energy that is equivalent to the power in kw times the number of hours the device worked.
In this case, it would be equal to:
1 kw also means 1kj of energy spent per second. With this, we calculate the amount of energy in kJ spent by the resistance:
Answer:
a) The magnitude of the magnetic field = 7.1 mT
b) The direction of the magnetic field is the +z direction.
Explanation:
The force, F on a current carrying wire of current I, and length, L, that passes through a magnetic field B at an angle θ to the flow of current is given by
F = (B)(I)(L) sin θ
F/L = (B)(I) sin θ
For this question,
(F/L) = 0.113 N/m
B = ?
I = 16.0 A
θ = 90°
0.113 = B × 16 × sin 90°
B = 0.113/16 = 0.0071 T = 7.1 mT
b) The direction of the magnetic field will be found using the right hand rule.
The right hand rule uses the first three fingers on the right hand (the thumb, the pointing finger and the middle finger) and it predicts correctly that for current carrying wires, the thumb is in the direction the wire is pushed (direction of the force; -y direction), the pointing finger is in the direction the current is flowing (+x direction), and the middle finger is in the direction of the magnetic field (hence, +z direction).
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
Answer:
option d and b..............
