A wire segment 1.2 m long carries a current I = 3.5 A

[Curious314]

Homework Statement

A wire segment 1.2 m long carries a current I = 3.5 A, and is oriented as shown in the figure. The +x-axis points directly into the page. A uniform magnetic field of magnitude 0.50 T pointing toward the -x direction is present as shown. What is the magnetic force vector on the wire segment?

Homework Equations

F_m=Il x B

l=1.2m

h sin θ
h cos θ

The Attempt at a Solution

1.2 sin 30° = 0.6m
1.2 cos 30°= 1.04m

F_m= I | i j k |
| 0 1.4m 0.6m |
|0.5T 0 0 |


=3.5 (0.3 j + 0.7 k)
= 1.05 j + 2.45 k

so that's my attempt... but is totally wrong, my answer is not near any of the options. Does anyone knows what am I doing wrong??

Thank you!

 

[TSny]

Does 1.2 sin (30) give you the y or the z component of the length?

Also, the direction of the length vector should be in the direction of the current. So, think about the signs of the y and z components of the length vector as well as the sign of the x-component of B.

 

[Curious314]

the Y component i think... not?

got you on the direction! I missed that

 

[TSny]

the Y component i think... not?

Yes, the y-component. But did you then substitute it into the correct location in the determinant?

 

[Curious314]

Thank you sooo much! :d

 

[Curious314]

Hello! So this is my lastest attemp:

Since I is a current in opposite direction, is -I (thanks Tsny!)

i j k
0 -0.6 -1.03
0.5T 0 0So:

-3.5[(-0.6*0)-(-1.03*0)]i-[(0*0)-(0.5*-1.03)]j+[(0*0)-(-0.6*0.5)]k

=-1.80j-1.05k

This are the numbers, but not the signs... What am I missing?

Thanks!

 

[TSny]

What is the sign of the x-component of B?

 

[Curious3141]

Hello! So this is my lastest attemp:

Since I is a current in opposite direction, is -I (thanks Tsny!)

i j k
0 -0.6 -1.03
0.5T 0 0


So:

-3.5[(-0.6*0)-(-1.03*0)]i-[(0*0)-(0.5*-1.03)]j+[(0*0)-(-0.6*0.5)]k

=-1.80j-1.05k

This are the numbers, but not the signs... What am I missing?

Thanks!

If the direction vector for the length of the wire is already being reckoned correctly (with the negative signs on the correct components), the current can just be taken as positive. You should use +3.5A rather than -3.5A.

Also, your magnetic field strength should be -0.5i, since it's going in the negative x-direction (pointing directly out of the page).

I would suggest also that you save the simplification till the last step - otherwise you get rounding errors with the $\sqrt{3}$, for example.

You can check if your answer is in the correct orientation by using the right-hand rule.

BTW, interesting username.