Acopper rod 0.570 � long and with a mass 0.05900 �� is suspended from two thin wire. at right angle to the rod is a uniform magnetic field of 0.670 � pointing into the page. find (a) the direction and (b) magnitude of the electric current to levitate the copper rod’s gravitation force.

a) Left to right

b) 1.51 A

Explanation:

a)

The gravitational force on the rod due to its mass is in downward direction. hence to levitate the rod, the magnetic force on the rod must be in upward direction.

The magnetic field is inward to page and magnetic force must be upward. Using right hand rule, the current must be flowing from left to right.

Left to right

b)

L = length of the copper rod = 0.570 m

m = mass of the rod = 0.059 kg

B = magnitude of magnetic field in the region = 0.670 T

θ = Angle between the magnetic field and rod = 90

i = current flowing throw the rod = ?

The magnetic force on the rod balances the gravitational force on the rod. hence

Magnetic force = gravitational force

mg = i B L Sinθ

(0.059) (9.8) = i (0.670) (0.570) Sin90

i = 1.51 A

v=ir = 3x15 =45 v