Question

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.

Answer 1

Answer:

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)