A long straight wire carries a current of 50 A towards the top of page. A square
ID: 1391544 • Letter: A
Question
A long straight wire carries a current of 50 A towards the top of page. A square circuit loop ABCD, 2 cm on a side, is placed so that sides AD and BC are parallel to the wire. The loop is rigid. The loop is not moving initially. The wire forming the circuit loop has a resistance per length of 10 ?/m. There is a 9 volt battery in the circuit loop. There is a current in the circuit bloop.
a) Calculate the current in the circuit loop due to the battery.
b) Draw in ? ?s and ? ?s in the regions 1, 2, 3 and 4 due to the magnetic field generated by the long wire. Try to match your drawing to the magnetic field strength in the region.
c) Determine the directions of the magnetic force on each side of the circuit loop (sides AB, BC, CD, and DA) due to the wire. Indicate these directions by drawing in the force vectors for each side in the above diagram. Do not worry about the presence of the battery.
d) Calculate the magnitude and direction of the net magnetic force on the circuit loop by the straight wire.
e) If the circuit loop was not fixed in place what do you think would happen to it and why? Answer in sentences.
f) If the circuit loop was fixed in place but was not rigid what do you think would happen to it and why? Answer in sentences and an illustration.
g) Calculate the magnitude and direction of the magnetic field at the center of the circuit loop due to the current through the circuit loop?s four sides.
h) Neglect the magnetic field due to the circuit loop and assume the strength of the magnetic field due to the wire at the center of the loop is an estimate of the average magnetic field through the loop due to the wire. Calculate the magnetic flux due to the wire through the loop.
i) If the current in the wire linearly drops to zero in 2 sec, what is the magnitude of the ?induced developed (and superimposed upon that due to the 9 V battery) in the loop during this 2 sec interval?
j) Indicate the direction of the current developed in the loop due to the current in the wire dropping as described in the above question.
Explanation / Answer
a) The current is:
V = IR ->I= V/R where
R = ?L so:
I = V/?L using given data we have:
I = 11.25 A
b) Regions
Region 1: .
Region 2: x
Region 3: x
Region 4: x
c) Direction of the force
left wire: To the left
top wire: upward
right wire: to the right
bottom wire: downward
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