Question

A wire with a mass of 1.14 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wire carries a current of 1.41 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?

Answer 1

Answer:
`B=0.0162T`

Step-by-step explanation:

let this sign be ∅ (titha in degrees)

let coefficient of friction of 0.200 be u

and i be current i= 1.41A

mass m=1.14g/cm m=0.0114g/m

To find the magnetic field from the north will be:

tan ∅= u

∅=
`tan^(-1)(0.200`

∅=11.3°

The direction:

i L*B cos(∅)= umg

divide both by iLcos(∅) to find B

B=
`(umg)/(iBcos(∅))` where m=
`(m)/(L)`

B=
`(u(m/L)g)/(icos(∅))`

B=
`(0.200*0.0114*9.81)/(1.41*cos(11.3))`

`B=0.0162T`