A square loop of wire of side 0.20 m has a total resistance of 0.3 Ohm. The loop
ID: 1613304 • Letter: A
Question
A square loop of wire of side 0.20 m has a total resistance of 0.3 Ohm. The loop is positioned in a uniform magnetic. field B of 7.5 T. The field directed into the page, perpendicular to the plane of the loop, as shown to the right. (a) Calculate the magnetic flux phi through the loop. The square loop is now pulled from the left and right so that it makes a rectangle that is 0.30 m times 0.10 m on. While the loop is being stretched an emf of 0.5 volts is induced. (b) Determine the length of time the loop was being stretched from a square to a rectangle. (c) i. Calculate the magnitude I of the ament in the loop during this period. ii. Determine the direction of the induced current while the loop is being stretched? ___ Clockwise ___ Counterclockwise The rectangular loop is now routed 90 degrees in the same time as determined in part (b), so that the plane of the loop is parallel to the magnetic field. (d) During this time the magnitude of the induced current is ____ when the loop is being stretched. ___ Greater Than ____ Less Than Justify your answer.Explanation / Answer
Given
square loop of length l = 0.2 m , area is A = 0.2^2 = 0.04 m2
resistance R = 0.3 ohm
magnetic field B = 7.5 T , directed in to page
a) magnetic flux through the loop is Phi = B*A cos theta
phi = 7.5*0.04 cos0 = 0.3Wb
b) when the square loop is stretched so that it became the rectangle with dimensions 0.3m X 0.1 m
area of the loop inside the magnetic field is A = 0.3*0.1 = 0.03 m2
the induced emf is e = 0.5 V
we know that from the motional emf induced emf e = -d phi /dt
dt = -d phi /e = -B*dA/e = - 7.5*(0.03-0.04)/0.5 = 0.15 s
c) i)
induced current in the loop is I = e/R = 0.5/0.3 = 1.667 A
ii) direction of the induced is counterclock wise direction
if the plane of the loop is in parallel to the magnetic field
the magnetic flux is zero because cos 90 = 0
the induced current is equal to zero
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