How would you determine the magnitude of the magnetic force on each of the four sides?

A square coil of wire containing a single turn is placed in a uniform 0.20 T0.20T magnetic field, as the drawing shows. Each side has a length of 0.32 m0.32m, and the current in the coil is 12 A12A.
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1 Answer
Mar 7, 2016

Upper and lower sides |vecF|=0.768NF=0.768N
Both vertical sides |vecF|=0F=0

Explanation:

Lorentz Force equation describes the force experienced by a charge particle having charge qq when it moves with velocity vecvv in an external electric field vecEE and magnetic field vecBB

vecF= q(vecE+ vecv times vecB)F=q(E+v×B)

In the problem it is given that there is there is no electric field.

Also, it is a case of wire carrying an electric current II placed in a magnetic field. Here each of the moving charges, which constitute the electric current in the wire, experiences the Lorentz force. The relation is:

vecF= I cdotvecl times vecBF=Il×B ,
where vecll has magnitude equal to the length of wire, and direction along the wire, and in the direction of current II.

For horizontal sides of the coil:

|vecF|=12cdot0.32cdot0.20sin theta=0.768NF=120.320.20sinθ=0.768N
:. sin 90^@=1

For vertical sides, cross product of vectors vanishes as sin theta=sin0^@=0. Therefore force is zero.

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For sake of completeness.

Direction of the force is given by the cross product of the two vectors. It can also be found with the help of right hand rule as in the picture below.
![wikimedia.org/wikipedia](useruploads.socratic.org)

Direction of the force experienced by upper side is out of the paper. Whereas for the lower side it is in to the paper.