CHAPTER 23 (18) * metal rod of muss m slides without friction along two parallel horizontal VAo rails separated by a distance l and connected 'by Shown in Fig Pz3.15. A'uniform vertical magnetic field of magnitude Bis applied perpendicular to the plane of the paper. The applied' force shown in fiqure acts of m, l, R, B, and 'v, find the distance the rod will then slide as it coasts to a stop. a resistor R, as for a' moment, to give the rod a speed v. In terms if unduction bar of length I moves

Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter10: Motion In A Noninertial Reference Frame
Section: Chapter Questions
Problem 10.13P
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CHAPTER 23 (18)
A metal rod of mass m slides without friction along two parallel horizontal
UAs rails, separated by a distance l and connected
magnitude B'is
applied perpendicular to the plane of the paper. The applied' force shown in
the fiqure acts only for a moment, to give the rod a speed v. In terms
of m, l,R, B, and 'v, find the distance the rod will then slide as it
coasts to a stop-
* if a conductiny bar of length l moves
through a onugntic field with a speed
v So that B is perpendicular to the ber,
the emf induud in the bar, often called
notional emf, is 1= Blv
x xTx x
X x x X
Fapp
イメ ×× ×
X xx x
X XX X
F= ma
E= BRV
「av: - at . -8父
IlB = ma
dt
EBl = mdv .dt
R
dt
v(e)
In vl.
-6と
dt E Bl = mdv
R
- In vle)- Inv - -B'l (ヒ-0)
-dtを Bl = dv
mR
RM:
: In v (t)
- dと BXv-dv
%3D
Rm
v(t)= ve
t
InR
For v(E)=0 t= o0
D =
ve
dt
Transcribed Image Text:CHAPTER 23 (18) A metal rod of mass m slides without friction along two parallel horizontal UAs rails, separated by a distance l and connected magnitude B'is applied perpendicular to the plane of the paper. The applied' force shown in the fiqure acts only for a moment, to give the rod a speed v. In terms of m, l,R, B, and 'v, find the distance the rod will then slide as it coasts to a stop- * if a conductiny bar of length l moves through a onugntic field with a speed v So that B is perpendicular to the ber, the emf induud in the bar, often called notional emf, is 1= Blv x xTx x X x x X Fapp イメ ×× × X xx x X XX X F= ma E= BRV 「av: - at . -8父 IlB = ma dt EBl = mdv .dt R dt v(e) In vl. -6と dt E Bl = mdv R - In vle)- Inv - -B'l (ヒ-0) -dtを Bl = dv mR RM: : In v (t) - dと BXv-dv %3D Rm v(t)= ve t InR For v(E)=0 t= o0 D = ve dt
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