Question

A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the z axis from z = 0 to z = 3.0m. In this region of space, the magnetic field is equal to 40 mT in the positive z direction. What is the magnitude of the force on this segment of wire?

Answer 1

`F_x=2N`

Step-by-step explanation:

The force on this segment is given by:

`F=I*(\beta x L)`

In the axis z'

`F_z=I*(\beta x L_z)`

`\beta x L_z=\beta *L*sin(0)=40mT*3.0m*0=0`

So the force is zero in the axis z' because the normal is in the same axis of the magnetic field

In the axis x'

`F_x=I*(\beta x L_x)`

`\beta x L_x=\beta *L*sin(90)=40mT*-2.0m*1`

`F_x=25A*80x10^(-3)T*m`

`F_x=2N`