.3x cwrnt (out of page 3. The is a positive const Loop \"A\" counterclockwise th
ID: 2305896 • Letter: #
Question
.3x cwrnt (out of page 3. The is a positive const Loop "A" counterclockwise the page, ge in the +z direction and is given by Beax, where a ant Iwo square current loops having sides of length L are placed in this field viewed from above, as shown. Loop "B" is in the yz-plane (perpendicular to magnetic field points out of the pa is in the xy-plane, with it's left side a distance x from the y-axis, and its current is and to loop "A"), with all of its sides a distance 3x from the y-axis, and its current such that it flows into the page at the bottom and out of the page at the top as shown. (a) Is there a net force on Loop A? If so, circle the direction RIGHT LEFT UP DOWN INTO PAPER OUT OF PAPER IS ZERO (b) Is there a net torque on Loop A, with respect to its center? If so, circle the direction: RIGHT LEFT UP DOwN INTO PAPER OUT OF PAPER IS ZERO (c) If either the force or torque or both is non-zero, calculate its magnitude(s) of any non-zero ones here, by the method of your choice: (d) Is there a net force on Loop B? If so, circle the direction: RIGHT LEFT UP DOWN INTO PAPER OUT OF PAPER IS ZERO (e) Is there a net torque on the Loop B? If so, circle the direction: RIGHT LEFT UP DOWN INTO PAPER OUT OF PAPER IS ZERO (D If either the force or torque or both is non-zero, calculate the magnitude(s) of any non-zero ones here, by the method of your choice:Explanation / Answer
3. given magnetic field points out of the page, in +z direction
given by B = alpha(x)
where alpha is a constant, greater than 0
loop side = L
Loop A in xy plane
left side distance x from y axis
clockwise counter clockwise current,
Loop B in yz plane
counterclockwise current
distnace from y axis = 3x
current out of the page from top, into the bage at bottom of the loop
a. loop A
the top and bottom sides, from symmetry will have same and opposing force acting on the loop
the left side will have less force acting to left, where as right side will have more force to the right as magnetic field is proportional to distance form the y axis and the force is proportional to the magnetic field
hence net force on loop A is to the right
b. as all the forces are balanced and are coplanar
net torque on the loop is 0
c. force on the loop = F
then
dF = [B(x + L) - B(x)]*i*dL ( where i is current in the loop)
hence
net force = integrating the equation = alpha(x + L - x)*i*L = alpha*L^2*i
d. as the loop B is in a constant magnetic field ( as magnetic field does not change in yz plane, it changes along x axis), hence net force on the loop B is 0
e. as the forces are not coplanat, the net torque on the loop B is non zero
now since the curernt comes out of the plane at top, hence the top edge has force to the left and bottom to the right
hence hte torque is along -ve y axis
f. value of this torque = (alpha*3x*i*L^2)
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