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: $B=0.0162T$

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= $\frac{umg}{iBcos(∅)}$  where m=$\frac{m}{L}$

B= $\frac{u(m/L)g}{icos(∅)}$

B= $\frac{0.200*0.0114*9.81}{1.41*cos(11.3)}$

$B=0.0162T$