Question

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 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 none of the above
NEED BOTH PARTS

Answer 1

a] Force on a current carrying wire due to a magnetic field is given by: $F = BILsin\theta$

where $\theta$ is the angle between the wire and the external magnetic field.

for the first part: $\theta$ = 0^o and so F = BILsin(0) = 0 N

b] $\theta$ = 90^o

6 = B(8.7)(0.35)sin90^o

=> B = 1.97 T in the negative z direction.