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A wire with a mass of 1.50 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wire carries a current of 1.52 A eastward and moves horizontally to the north. What are the magnitude and the direction of the smallest magnetic field that enables the wire to move in this fashion?

Respuesta :

Answer:

0.193 T

Inward direction

Explanation:

[tex]\frac{m}{L}[/tex] = Mass per unit length of wire = 1.50 g/cm = [tex]\frac{1.50\times 10^{-3}kg}{0.01 m}[/tex] = 0.15 kg/m

[tex]i[/tex] = magnitude of current = 1.52 A

[tex]B[/tex] = magnitude of magnetic field = ?

[tex]m[/tex] = mass of the wire

[tex]L[/tex] = length of the wire

μ = Coefficient of friction = 0.200

For the wire to move,

magnetic force = frictional force

[tex]i[/tex] [tex]B[/tex] [tex]L[/tex] = μ [tex]m[/tex] [tex]g[/tex]

(1.52) [tex]B[/tex] [tex]L[/tex] = (0.200) (9.8) [tex]m[/tex]

(1.52)  [tex]B[/tex] = (0.200) (9.8) [tex]\frac{m}{L}[/tex]

(1.52)  [tex]B[/tex] = (0.200) (9.8) (0.15)

[tex]B[/tex] = 0.193 T

Direction of magnetic force is towards north and current is directed towards east. hence using right hand rule, the direction of magnetic field comes out to be in inward direction.

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