A straight, vertical wire carries a current of 2.80 A downward in a region betwe

Question

A straight, vertical wire carries a current of 2.80 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude B = 0.513 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field in the following scenarios?

(a) the magnetic field direction is east
N

(b) the magnetic field direction is south
N   
(c) the magnetic field direction is 30.0° south of west
N

(d) Direction of A,B,C?

Explanation / Answer

Force on current carrying wire in a magnetic field

F = i*(LxB) = BILsin(theta)

a) F = 0.513*2.80*0.01*sin(90 degrees) = 0.01436 N outward direction

b) F =  0.513*2.80*0.01*sin(0 degrees) = 0 N  

c) F =  0.513*2.80*0.01*sin(60 degrees) = 0.0124 N inward direction.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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