Awire 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?

0.193 T

Inward direction

Explanation:

= Mass per unit length of wire = 1.50 g/cm = = 0.15 kg/m

= magnitude of current = 1.52 A

= magnitude of magnetic field = ?

= mass of the wire

= length of the wire

μ = Coefficient of friction = 0.200

For the wire to move,

magnetic force = frictional force

= μ

(1.52) = (0.200) (9.8)

(1.52)   = (0.200) (9.8)

(1.52)   = (0.200) (9.8) (0.15)

= 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.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s