A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from
1 answer:
Answer:
The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.
Explanation:
Given;
electric current, I = 21.3 A
distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m
The strength of the resulting magnetic field at the given distance is calculated as;
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A
Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.
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Answer:
The magnitude of the magnetic force acting on the conductor is 0.75 Newton
Explanation:
The parameters given in the question are;
The length of the straight segment of wire, L = 25 cm = 0.25 m
The current carried in the wire, I = 5 A
The orientation of the wire with the magnetic field = Perpendicular
The strength of the magnetic field in which the wire is located, B = 0.60 T
The magnetic force, 'F', is given by the following formula;
F = ·L×
= I·L·B·sin(θ)
Where;
= The current flowing, I
L = The length of the wire
= The magnetic field strength, B
θ = The angle of inclination of the conductor to the magnetic field
Where I = 5 A, L = 0.25 m, B = 0.60 T, and θ = 90°, we get;
F = 5 A × 0.25 m × 0.60 T × sin(90°) = 0.75 N
Therefore
The magnitude of the magnetic force, F = 0.75 N.