Respuesta :
Answer:
F = 0.0034 N
Explanation:
Given:
[tex]B = 5.2*10^(-5) T\\Q = 57 degrees\\I_{wire} = 12 A\\L_{wire} = 10 m[/tex]
The angle between B and wire = 90 - 57 = 33 degrees
Using formula:
[tex]F = B*I*L*sin (90-Q)\\F = (5.2*10^(-5)*(12)*(10)*sin (33)\\F = 0.0034 N[/tex]
(A) The magnetic force exerted on the wire is 3.4×10⁻³N
(B) The direction of the force is to the west.
Magnetic force:
Given that the magnetic field B = [tex]5.2\times10^{-5}T[/tex] which points in the direction 57° below a horizontal line pointing north.
Length of the wire L = 11m
current in the wire I = 11A
The angle between the wire and the magnetic field is θ = (90-57) = 33°
(A) The magnetic force on a finite wire of length L carrying a current I is given by:
[tex]F=BILsin\theta\\\\F=5.2\times10^{-5}\times11\times11\sin33\\\\F=3.4\times10^{-3}N[/tex]
(B) The direction of the force is given by dl×B, now B is at 57° with the north direction and the wire is verticle, so the direction of the field will be to the west.
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