# 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 $0.20 T$ magnetic field, as the drawing shows. Each side has a length of $0.32 m$, and the current in the coil is $12 A$. Mar 7, 2016

Upper and lower sides $| \vec{F} | = 0.768 N$
Both vertical sides $| \vec{F} | = 0$

#### Explanation:

Lorentz Force equation describes the force experienced by a charge particle having charge $q$ when it moves with velocity $\vec{v}$ in an external electric field $\vec{E}$ and magnetic field $\vec{B}$

$\vec{F} = q \left(\vec{E} + \vec{v} \times \vec{B}\right)$

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 $I$ 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:

$\vec{F} = I \cdot \vec{l} \times \vec{B}$ ,
where $\vec{l}$ has magnitude equal to the length of wire, and direction along the wire, and in the direction of current $I$.

For horizontal sides of the coil:

$| \vec{F} | = 12 \cdot 0.32 \cdot 0.20 \sin \theta = 0.768 N$
$\therefore \sin {90}^{\circ} = 1$

For vertical sides, cross product of vectors vanishes as $\sin \theta = \sin {0}^{\circ} = 0$. 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. Direction of the force experienced by upper side is out of the paper. Whereas for the lower side it is in to the paper.