A magnetic field is used to suspend a wire of mass 0.04 kg and length 0.13 m . The wire is carrying a current of 15 A . What minimum magnetic-field magnitude is needed to balance the pull of gravity

0.087 T

Explanation:

Note: The weight of the mass suspended is equal the force produced by the current  carrying conductor in the magnetic field and this is the force needed to balance the pull of gravity

Fm = mg = BIL Equation 1

Make B the subject of the equation

B = mg/IL Equation 2

Where B = magnetic Field, m = mass, g = acceleration due to gravity, I = Current, L = Length of the wire.

Given: m = 0.04 kg, I = 15 A, L = 0.13 m,

Constant: g = 9.8 m/s²

Substitute into equation 2

B = (0.04×9.8)/(15×0.13)

B = 0.392/4.5

B = 0.087 T

it takes about 30 days for it to break down

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6