The 10-turn of loop of wire shown lies in a horizontal plane, parallel to a unif
ID: 1501674 • Letter: T
Question
The 10-turn of loop of wire shown lies in a horizontal plane, parallel to a uniform horizontal magnetic field, and carries a 2.0 A current. The loop is free to rotate about an axis through the center (see diagram). A 50 g mass hangs from one edge of the loop. What magnetic field strength will prevent the loop from about the axis? (2. 10 T) A cross section of three long wires with linear mass density of 50g/m us shown. They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table. What current will allow the upper wire to float so as to form an equilateral triangle with the lower wires? (238 A)Explanation / Answer
1). Relevant equations:
F = ILxB
torque = FR
2). The attempt at a solution:
First I attempt to sort out the torques to see what cancels what. Since the wire loops are suppose to stay parallel to the magnetic field, the sum of the torques should be zero. Another thing to note is that the magnetic field only acts on the wire where they go perpendicular to the field. Since there are 10 loops here, that means that we have 10 times the force on each side of the wire loop
But torque = Number of turns(mxB) = 10 x IA x B
So mgr-10IAB=0
mgr-10IABr=0
mgr=10IAB
(mg)/(10IL) = B
(.05*9.81)/(10*2*.1) = 2.1 Tesla
From what I understand, its saying that it's a coil or stack of ten loops, each loop adds additional torque and reduces the magnetic field strength needed to create an overall torque
2)
F=uLI^2/(2pid)
F=mg
since current in lower wires are in opposite direction from top wires, therefore they repel each other.
The y-components from each lower two wires must equal force of gravity of top wire:
mg = 2 x uLI^2/(2pid)sin30
solve for I, assume L=1 m and where d=0.04m, mg=0.4905 N
you get , I=238A
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