A stiff wire 50.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y=2x in the xy plane. A current of 20.0 A flows in the wire --- down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field B = (.318 i)T. Determine the magnitude and the direction of the total force on the wire.

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

Explanation:

The stiff wire 50.0cm long bent at a right angle in the middle

One section lies along the z axis and the other is along the line y=2x in the xy plane

tan θ = 2

Therefore,

slope m = tan θ = y / x

Then length of each section is 25.0cm

so, length vector of the wire is

And magnetic field is B = (0.318T)i

Therefore,

Magnitude of the force is

Direction is

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

F=2.38i - 4.57j

magnitude of F=5.15N

Explanation:

The force on the wire is given by

in this problem we have to compute the force for both sections if the wire. Hence we have

The direction of l1 is k. The direction of l2 is obtained by using the slope of the line y=2x

By applying the cross product we have

and its magnitude is

HOPE THIS HELPS!!