Given Information:
Current = I = 28 A
distance between wires = r = 2.0 m
Required Information:
Magnetic field = B = ?
Answer:
B = 12x10⁻⁶ T
Step-by-step explanation:
Biot-Savart Law is given by
B = μ₀I/2πr
Where μ₀ is the permeability of free space, I is the current flowing through the wire and B is the magnitude of the magnetic field produced.
We are asked to find the magnetic field midway between the wires so r/2 = 1
B = 4πx10⁻⁷*28/2π*1
B = 6x10⁻⁶ T
since the same amount of current flows in both wires therefore, equal amount of magnetic field will be produced in both wires
B = 2*6x10⁻⁶ T
B = 12x10⁻⁶ T
Therefore, the net magnetic field midway between the two wires is 12x10⁻⁶ T.
Answer:
The tensile stress on the wire is 550 MPa.
Explanation:
Given;
Radius of copper wire, R = 3.5 mm
extension of the copper wire, e = 5.0×10⁻³ L
L is the original length of the copper wire,
Young's modulus for copper, Y = 11×10¹⁰Pa.
Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.
Therefore, the tensile stress on the wire is 550 MPa.
Answer:
Explanation:
Check attachment for solution
The average weight of an athlete should be around 60kg so from the information that the athlete can run 100m in 10s, we can calculate that their average speed is 10m/s. Using the kinetic energy formula, Ek = 1/2mv^2 we can calculate the kinetic energy using 60kg as the mass.
(1/2)(60)(10^2) = Ek
Ek= 3000J
