# Force on current carrying square loop

I'm asked to find force on square loop (side a) carrying current $I$, flowing counter clockwise, when we look down x-axis, lying in yz plane. the loop is centered at the origin. The magnetic field is given as:

$\vec{B} = kz\hat{x}$

Its solution states that force on left an right cancel each other .The force on top is $IaB=iak(a/2)$ pointing upward and the force on bottom is$IaB=-iak(a/2)$ also pointing upward .How the force on bottom is upward? (From where minus sign came?). By R.H.R it should be downward.

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I'm not sure I understand how the loop is set but your case must be similar to the left and right side of the loop below.Since the current is flowing in a loop, on the left side it will flow the opposite direction of the way it does on the other side.Using the right-hand rule you will now see that the forces are equal and opposite,as shown in the picture, explaining the minus in your problem.

So, watch out which direction the current is going, always use the R.H.R. and carefully you will have no problem.

The minus is coming from the value of z. Note that the loop is centered at the origin on the yz plane, and the value of B is dependent on the value of z.

The forces on the left and right arms of the loop cancel out, the forces on the top and bottom arm are also opposite but due to the nature of the B field (dependence on z) the values are opposite in sign.

So one is $iak(a/2)$ in the $\hat z$ direction and the other is $iak(-a/2)$ in the $-\hat z$ direction. so when you add them together, instead of cancelling out you are left with double of the force.

first you take the direction of current vector and start turning /rotating screw from I towards the B vector ;if the rotation of the screw is clockwise screw movement will be perpendicular to the plane containing Current and Field vector and it will give you direction of the force . if the screw rotates anticlockwise it is coming out backwards this direction will be direction of the force- naturally on your rectangular frame the two arms carrying currents which are perpendicular/or making an angle with the B field will experience a couple providing a torque let us see if it works! for those currents which are in the direction or opposed to magnetic field direction will not experience any forces as the vector product will vanish.

You are correct, and the "solution" is apparently in error. The force on the top leg is equal and opposite to the force on the bottom leg. They cancel just as the forces on the left and right legs do. There is no net force on the loop.

This problem is given in griffith book of electrodynamics. Here minus sign comes from the negative z direction because loop is placed at centre half of which in positive axis and remaining half lies in negative axis.