NEED THIS DONE ASAP HELP
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Answer:
All of the above
Explanation:
The magnitude of the magnetic force on a current-carrying wire held in a magnetic is given by the equation
Where B = Strength of the magnetic field
I = The current carried by the wire
l = length of the wire in the magnetic field
θ = Angle between the wire and the magnetic field
Based on the relationship written above, the magnitude of the magnetic force on the current - carrying wire in the magnetic field depends on the strength of the magnetic field (B), length of the wire(l), current in the wire (I).
All the options are correct.
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S = 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S > σ