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)
