Question

The magnetic field in a region of space is specified by its x, y, and z components which are respectively B_x = 0.138 T, B_y = 0.162 T, and B_z = 0.200 T. If a 30.0-cm wire carrying a current of 3.00 A is oriented along the e-axis with the current in the -z direction, determine the magnitude and direction of the force acting on the wire. magnitude N direction degree counterclockwise from the +x-axis

Answer 1

Given the magnetic field Bx = 0.138 T, By = 0.162 T, Bz = 0.200

length of the wire L = 0.3 m, curent in the -ve z direction is A

we know that the force F = I lB sin theta

           F = lI X B
cross product of I,B is

i j k

0 0 -3

0.138   0.162 0.200

i(0 + 3*0.162) -j (0+3*0.138)+k(0-0)

F = 0.3((0.486) i - 0.414 j + 0k)

F = 0.1458 i - 0.1242 j +0k

magnitude is F = sqrt((0.1458)^2+(-0.1242)^2) N = 0.1915287967905 N = 0.192 N

the direction is theta = arc tan (-0.1242/0.1458) = -40.426 degrees