enter image description here

Please assume that the conductor of wire exists , and the current flows from below left to above right like shown in the above diagram.

As shown in the above diagram , the uniform magnetic fields are given .

I want to know the forces which act against the wire.

I tried to use the rule of Flemming left hand .

But this rule seemingly only be able to be used as an angle between a magnetic field and a current , is a right angle .

I know the below formula .

$$ \boldsymbol{f}= l \left( \boldsymbol{I}\times\boldsymbol{B} \right) $$

Hence, the below is held .

$$ \Vert \boldsymbol{f} \Vert = l \cdot \left(\Vert \boldsymbol{I} \Vert \cdot \Vert \boldsymbol{B} \Vert \cdot \sin^{}\left(\theta_{} \right) \right) $$

Just simply applying the above equation of magnitude can easily obtain the magnitude of total force which acts against the wire however how can I obtain the direction of the force vector?

I think the directions of it must be perpendicular against the directions of the magnetic fields but I don't know how to prove it so far.

I looked the page of wikipidea of cross product and found the below image .

enter image description here

Cited it from here

Moreover how can I determine the vectors of forces as the directions of the magnetic fields are reversed?

  • \$\begingroup\$ Why would you change to the second form of the equation and not stick with the vector form? Just use the vector form and all the magnitudes and directions pop out. Do you not understand how to directly work with dot and cross product where you don't convert it to a form that contains a cos or sin? \$\endgroup\$
    – DKNguyen
    Commented Aug 23, 2021 at 3:57
  • \$\begingroup\$ efcms.engr.utk.edu/ef151-2019-08/m4/class-4-5/img/… \$\endgroup\$
    – DKNguyen
    Commented Aug 23, 2021 at 4:07
  • \$\begingroup\$ Yes , I think I don't understand it so far. \$\endgroup\$ Commented Aug 23, 2021 at 4:48
  • \$\begingroup\$ Yeah, I don't think you do. The conversion from vector for to magnitude form with trig functions is not the way the vectors are meant to be used. Leave it cross product form and treat it like a multiplication WHERE ORDERS MATTERS. treat i,j,k as variables (they are unit vectors with magnitude of 1 so do not change the magnitude when multiplied but have a direction associated with them). \$\endgroup\$
    – DKNguyen
    Commented Aug 23, 2021 at 4:50
  • \$\begingroup\$ The order the multiplication happens matters according to the image I linked and when you multiply (cross) two unit vectors together, it turns into the third unit vector with an associated sign depending on the order. \$\endgroup\$
    – DKNguyen
    Commented Aug 23, 2021 at 4:54

1 Answer 1


$$ \boldsymbol{ F }_{} = \boldsymbol{I}\times\boldsymbol{B} $$

$$ d \boldsymbol{ F }_{} = I \left( d \boldsymbol{ s }_{} \times \boldsymbol{ B }_{} \right) $$

Use these with right hand rule .


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