Question

A wire with a mass of 1.50 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wire carries a current of 1.52 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:

0.193 T

Inward direction

Explanation:

$\frac{m}{L}$ = Mass per unit length of wire = 1.50 g/cm = $\frac{1.50\times 10^{-3}kg}{0.01 m}$ = 0.15 kg/m

$i$ = magnitude of current = 1.52 A

$B$ = magnitude of magnetic field = ?

$m$ = mass of the wire

$L$ = length of the wire

μ = Coefficient of friction = 0.200

For the wire to move,

magnetic force = frictional force

$i$ $B$ $L$ = μ $m$ $g$

(1.52) $B$ $L$ = (0.200) (9.8) $m$

(1.52)  $B$ = (0.200) (9.8) $\frac{m}{L}$

(1.52)  $B$ = (0.200) (9.8) (0.15)

$B$ = 0.193 T

Direction of magnetic force is towards north and current is directed towards east. hence using right hand rule, the direction of magnetic field comes out to be in inward direction.