A wire of length 0.35 m is conducting a current of I = 8.7 A in the +x direction

Question

A wire of length 0.35 m is conducting a current of I = 8.7 A in the +x direction through a B = 5.0-T uniform magnetic field that is directed parallel to the wire in the -x direction. What are the magnitude and direction of the magnetic force on the wire? +x -x +y -y +z -z none of the above The magnetic field changes, so that now the force on the wire is F = 6.0 N in the +y direction. What are the magnitude and direction of the magnetic field now? (Assume that the magnetic field B is perpendicular to I.) +x -x +y -y +z -z

Explanation / Answer

given: 0.35m=l
I=8.7A
B=5.0T

angle=90deg

magnitude and direction of the magnetic force

F=BIL sin(angle)

F=5*8.7*0.35*sin(90)=15.22N

direction left

________________________________________

force = ILx^ X B

6 y^/IL = x^ X B

now from here direction of B is (-Z). for magnitude value of currrent and wire length is required.

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