A horizontally oriented coil of wire of radius 5 cm and carrying a current, i, i

Question

A horizontally oriented coil of wire of radius 5 cm and carrying a current, i, is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the coil. If the magnetic field on all parts of the coil makes an angle of 450 with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in midair. The magnitude of the magnetic field is B = 0.010 T, the number of turns in the coil is N = 10, and the total coil mass is 10 g

Explanation / Answer

F =NBIAsin

L = 2r=2*3.14*0.05=0.314m

the force should be equal to weight of coil.

so F =mg =0.01*9.8=0.098N

so

0.098=10*.01*I*0.314*sin(45)

I = 0.098/(0.0314*0.707) = 4.144 A

so current should be equal to 4.144 A.

please rate me lifesaver.

thanks.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.