Problem 2: A square loop, length L on each side, is shot with velocity vo into a uniform magnetic field B. The field is perpendicular to the plane of the loop (into the page in Fig. 2). The loop has mass m and resistance R, and it enters the field at t = 0 s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0 m. Find an expression for the loop's velocity as a function of time as it enters the magnetic field.

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Chapter13: Electromagnetic Induction
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Problem 2: A square loop, length L on each side, is shot with velocity
vº into a uniform magnetic field B. The field is perpendicular to the
plane of the loop (into the page in Fig. 2). The loop has mass m and
resistance R, and it enters the field at t = 0 s. Assume that the loop
is moving to the right along the x-axis and that the field begins at
x = 0 m. Find an expression for the loop's velocity as a function of
time as it enters the magnetic field.
X
X X
L
X
GIV
a) The induced emf is & = dem, where the change of the flux
dt
Pm is due to the change of the area of the loop inside the magnetic
field. At the moment of time shown in the scheme this area is equal to xL. Derive the formula for the
induced current and express it in terms of the loop's speed v = d. Is the current flowing clockwise or
counterclockwise?
FIG. 2: The scheme for Problem 2
b) Express the magnetic force that is acting on the loop in terms of B, L, R, and v. Notice that the net
force is the force acting on the right edge of the loop, because the forces acting on its top and bottom edges
(their parts inside the region with the field) cancel each other. Is the force in the positive or negative î
direction?
c) Write down the Newton's second law, ma = F, for the loop (use only the magnetic force and ignore
the gravitational force). Show that this equation can be reduced to du Bdt (where does the minus
sign come from?).
mR
d) Integrate the left side of this equation with respect to v and the right side with respect to t from the
initial state (t = 0, v = vo, when the loop enters the field) to the intermediate state (t, v). Assume that the
left edge of the loop has not entered the field by the time t. The result will be the expression for v as a
function of t.
Transcribed Image Text:Problem 2: A square loop, length L on each side, is shot with velocity vº into a uniform magnetic field B. The field is perpendicular to the plane of the loop (into the page in Fig. 2). The loop has mass m and resistance R, and it enters the field at t = 0 s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0 m. Find an expression for the loop's velocity as a function of time as it enters the magnetic field. X X X L X GIV a) The induced emf is & = dem, where the change of the flux dt Pm is due to the change of the area of the loop inside the magnetic field. At the moment of time shown in the scheme this area is equal to xL. Derive the formula for the induced current and express it in terms of the loop's speed v = d. Is the current flowing clockwise or counterclockwise? FIG. 2: The scheme for Problem 2 b) Express the magnetic force that is acting on the loop in terms of B, L, R, and v. Notice that the net force is the force acting on the right edge of the loop, because the forces acting on its top and bottom edges (their parts inside the region with the field) cancel each other. Is the force in the positive or negative î direction? c) Write down the Newton's second law, ma = F, for the loop (use only the magnetic force and ignore the gravitational force). Show that this equation can be reduced to du Bdt (where does the minus sign come from?). mR d) Integrate the left side of this equation with respect to v and the right side with respect to t from the initial state (t = 0, v = vo, when the loop enters the field) to the intermediate state (t, v). Assume that the left edge of the loop has not entered the field by the time t. The result will be the expression for v as a function of t.
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