A wire with a linear mass density of 1.19 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic friction between the wire and the surface is 0.250. If the wire carries a current of 1.76 A westward and moves horizontally to the south, determine the magnitude and direction of the smallest magnetic field that can accomplish this.

The magnitude and direction of the smallest magnetic field that can accomplish this are 0.172 T and vertically to the east.

Explanation:

given information:

density of wire, m/l = 1.19 g/cm = 0.119 kg/m

coefficient of kinetic friction, μk = 0.250

current, I = 1.7 A

according to the Newton's first law

∑F = 0

F(magnetic) - F(friction) = 0

F(magnetic) = F(friction)

F(magnetic) = B i l sin θ

where

F (magnetic) = magnetic force

B = magnetic field (T)

i = current (A)

l = the length of wire (m)

θ = angle

and

F(friction) = μk N

where

F(friction) = kinetic force (N)

μk = coefficient of kinetic friction

N = Normal force (N)

so,

F(magnetic) = F(friction)

B i l sin θ = μk N

B = μk N/i l sin θ

now lets determine the normal force

N - mg = 0

N = mg

therefore

B = μk mg/i l sin θ

θ = 90° because the magnetic field is perpendicular to the wire (smallest magnetic field)

B = μk mg/i l sin 90

   = μk mg/i l

   = (μk g/i)  (m/l)

   = ((0.250) (9.8) / (1.7)) ((0.119)

    = 0.172 T

according to the right hand rule,  the direction of the magnetic field is vertically to the east. (magnetic force to the south and current to the westward)

i dont know but

explanation:

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