A square current loop lies flat in the x-y plane with its sides parallel to the
ID: 2012935 • Letter: A
Question
A square current loop lies flat in the x-y plane with its sides parallel to the x and y axes. The side of the square is a. The current flowing in the loop is I, in the CCW direction as viewed from above. A uniform magnetic field B = B(i/sqrt(2) + k/sqrt(2) is applied to the loop. (With B in the formula as Bo, i being i hat and k being k hat)a) Find the magnetic force (magnitude and direction) on each side of the loop. Check that the sum of the four forces equals zero.
b) Find the torque (magnitude and direction) on the current loop.
Explanation / Answer
Hope this helps This problem involves two formulas, one from relationship between the current and the magnetic field to the find the force by the magnetic field. The formula can be written as F=ILxB, where I is the current, L is the length of the wire where the magnetic field is applied, and B is the magnetic field. For problem part a), we can see that the formula applies simpler when the force of each side of the loop is calculated separately, like the instruction. Given that the length of the side is a, which we will use as L, and B is given by B0(ihatv2+jhatv2), and current as I, we can see that B has magnitude of 2B0, (by using distance formula of having ihatv2 as x displacement and jhatv2 as y displacement). Furthermore, we can see that since the x component and y component of the B vector is the same, we can deduce that the direction of the magnetic field is p/4 in terms of the x-y axes. Now, we can apply the formula for magnetic force. Since the formula is written in terms of cross product, it's easier to see the calculation in terms of magnitiude, which becomes F=I*L*B*sin? (All the variables here are written in magnitudes thus B here is the magnitude of B, which is 2B0) F=I*L*B*sin?=I*a*2B0*v2/2=v2*I*a*B0. Other sides will have the same magnitude since the sine of angle between the current and the magnetic field stays the same. (as p/4). In terms of direction, we can use the right hand rule, where we know that 2 sides (the sides that touch the origin) has force directed up (out of the page) and two sides directed down (into the page). And since the magnitude of the force for each side is the same, we can see that the net force on the loop is zero. b)In here, we can use the formula for torque, which is t=rxF, where r is the distance from the axis of rotation to the lever arm and the force applied. We know the axis of rotation of the piece since the two sides adjacent to the origin has outward force, and the other two inward. We can deduce that the axis or rotation as the diagonal that goes across the top left point and bottom right point of the loop. Since the rotation axis and the force applied has an angle of p/2 (the force applied is the cross product of two vectors in the same plane, ie. is perpendicular to the rotation axis which is also in the xy plane), we can apply for formula. However, we need to find r, since the force is applied throughout the wire, and the wire isn't a constant distance away from the axis. Because the force is distributed throughout the wire, we can isolate a segment of the wire, and call the force applied in the segment as dF. Thus, the formula will be t=rxdF for a small segment of wire, we can call as da. Since we know that the axis of rotation and the side always make p/4 angle in this situation we can calculate r in terms of a(temporarily used as a variable, but will get full a by integrating) which is r=a*sin?=v2/2*a. Thus t=integral of rxdF=integral of v2/2*a*d(v2*I*a*B0). (since it's integrated over small segment of a), =integral of a*I*B0 da =a^2/2*I*B0. But since a ranges over from 0 to a, a stays the same, and t=a^2/2*I*B0. However, since there are 4 segments that have the same magnitude, tnet=4*a^2/2*I*B0=2*a^2*I*B0. Thus the magnitude of the torque is , t=rxF= For the distance, we can use the right hand rule for the torque. Give that axis of rotation as the direction of the thumb, and the curling of the right fingers as the direction of rotation (which is more clockwise when viewed from x axis), we can see that the direction of the torque is towards top left point of the loop, ie 3p/4.
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