A counterclockwise current I is induced in the loop. The magnetic force FR on the bar carrying this current opposes the motion. Figure 30.8 (a) A conducting bar sliding with a velocity v along two conducting rails under the action of an applied force F Because of the current in the bar, there is a magnetic force F,on the bar in the direction opposite to the applied force. (b) The equivalent circuit diagram for the setup shown in (a). R F, app R |El = Blv app a I * * x x x B,n a 'in R Figure P30.18

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Chapter1: Units, Trigonometry. And Vectors
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You are working in a laboratory that uses motional emf to make magnetic measurements. You have found that it is difficult to create a uniform magnetic field across the entire sliding-bar apparatus shown in 30.8a, with a resistance R connected between the rails. You decide to investigate creating the magnetic field with a long, straight, current- carrying conductor lying next to and parallel to one of the rails, as shown in P30.18. This will create a nonuniform field across the plane of the bar and rails. You set up the apparatus in this way, with the current-carrying wire a distance a from the upper rail. You wish to find an expression for the force necessary to slide the bar at a constant speed of υ to the right in P30.18 if the wire carries a current I. (Hint: Two separate integrations will be required.)

A counterclockwise current I is
induced in the loop. The magnetic
force FR on the bar carrying this
current opposes the motion.
Figure 30.8 (a) A conducting
bar sliding with a velocity v along
two conducting rails under the
action of an applied force F
Because of the current in the bar,
there is a magnetic force F,on
the bar in the direction opposite
to the applied force. (b) The
equivalent circuit diagram for the
setup shown in (a).
R
F,
app
R
|El = Blv
app
a
Transcribed Image Text:A counterclockwise current I is induced in the loop. The magnetic force FR on the bar carrying this current opposes the motion. Figure 30.8 (a) A conducting bar sliding with a velocity v along two conducting rails under the action of an applied force F Because of the current in the bar, there is a magnetic force F,on the bar in the direction opposite to the applied force. (b) The equivalent circuit diagram for the setup shown in (a). R F, app R |El = Blv app a
I
* * x x x B,n
a
'in
R
Figure P30.18
Transcribed Image Text:I * * x x x B,n a 'in R Figure P30.18
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