Net magnetic force in unit-vector notation

[Mike88]

Homework Statement

The bent wire shown in the figure below lies in a uniform magnetic field. Each straight section is 2.0 m long and makes an angle of θ = 65° with the x axis, and the wire carries a current of 2.0 A.

(a) What is the net magnetic force on the wire in unit-vector notation if the magnetic field is 4.0k^{^}T?
magnitude [answer] N , direction [ ]

Homework Equations

F_B = iLBsin(phi)

The Attempt at a Solution

given i = 2A, L = 2m, B=4kT sin(θ) = 65° i got 2*2*4*sin(65) = 14.5 * 2 for other wire = 29N

 

[Doc Al]

Homework Equations

F_B = iLBsin(phi)

This gives you the magnitude of the force on the wire segment. Note that phi is the angle between iL and B. What direction is the B field?

What's the direction of the magnetic force on the wire segment? (Use the right hand rule.)

 

[thomate1]

The net magnetic force on the wire can be calculated using the formula FB = iLBsin(θ), where i is the current, L is the length of the wire, B is the magnitude of the magnetic field, and θ is the angle between the wire and the magnetic field.

In this case, the wire is bent and has two straight sections, each with a length of 2.0 m and making an angle of 65° with the x-axis. The wire carries a current of 2.0 A and is placed in a uniform magnetic field with a magnitude of 4.0k^T.

Using the given values, we can calculate the net magnetic force on the wire as follows:

FB = (2.0 A)(2.0 m)(4.0k^T)sin(65°) + (2.0 A)(2.0 m)(4.0k^T)sin(65°)
= (14.5 N)(2.0) + (14.5 N)(2.0)
= 29 N + 29 N
= 58 N

Therefore, the net magnetic force on the wire is 58 N, and its direction can be described using unit-vector notation as [58 N, 58 N, 0]. This indicates that the force is acting in the positive x and y directions, while there is no force in the z direction.