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In Houston, Earth's B→ field has a magnitude of 5.2 × 10−5 Tand points in a direction 57∘ below a horizontal line pointing north.Part ADetermine the magnitude of the magnetic force exerted by the magnetic field on a 11-m-long vertical wire carrying a 11-A current straight upward. Express your answer to two significant figures and include the appropriate units.Part BDetermine the direction of the magnetic force.a. to the northb. to the eastc. to the southd. to the west

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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