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

Answer:

$F_x=2N$

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'

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

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

$F_x=25A*80x10^{-3}T*m$

$F_x=2N$

Answer 2

The magnitude of the force on this segment of wire is mathematically given as

F_x=2N

What is the magnitude of the force on this segment of wire?

Question Parameter(s):

A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0

then along the z axis from z = 0 to z = 3.0m

Generally, the equation for the  Force is mathematically given as

F=I*(B x L)

Therefore

For z axis

Bx Lz=B *L*sin(0)

Fz=40mT*3.0m*0

Fz=0

For x axis

Fx=I*(B x Lx)

Fx=25A*80x10^{-3}

F_x=2N

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