A straight, horizontal length of copper wire has a current i=28A through it. What are the magnitude and direction of the minimum magnetic field B needed to suspend the wire(to balance the gravitational force on it) if the linear density of the wire is 46.6g/m?

Respuesta :

Answer:

0.01631 T

Explanation:

current, i = 28 A

mass per unit length, m/l = 46.6 g/m = 0.0466 kg/m

Let the magnetic field is B.

the weight of the wire is balanced by the magnetic force .

mg = i l B

B = mg / i l

B = (m/l) x g/i

B = 0.0466 x 9.8 / 28

B = 0.01631 T

Thus, the magnetic field is 0.01631 T.