*forgot in picture but I=2 A going this direction on the top wire I have a quest

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*forgot in picture but

 I=2 A going this direction on the top wire

I have a question on the following AP Physics B Free-response question, any help will be greatly appreciated. Please shown steps :D

Two parallel conducting rails, separated by a distance L of 2.0 m, are connected through a resistance of 3.0 as shown in the diagram. A uniform magnetic field with a magnitude B of 2.0 T points into the page. A conducting bar with mass m of 4.0 kilograms can slide without friction across the rails.

a. Determine at what speed the bar must be moved, and in what direction, to induce a counterclockwise current I of 2.0 A as shown.

b. Determine the magnitude and direction of the external force that must be applied to the bar to keep it moving at this velocity.

c. Determine the rate at which heat is being produced in the resistor and determine the mechanical power being supplied to the bar.

d. Suppose the external force is suddenly removed from the bar. Determine the energy in joules dissipated in the resistor before the bar comes to rest.

Thanks for all help in advanced.

Explanation / Answer

a) E=Bvl (where E is emf)
using the right hand rule, to get the +ve charges to move up the rod, velocity must be directed towards the right.
E=iR
iR=BvL
v=iR/BL
=2*3/(2*2)
=1.5m/s

b) The external force has to balance the force cause by the induced emf

so F=Bil=2*2*2

   =8N

c) P=i2R=2*2*3

    =12J

mech Power=Fv =8*1.5

=12J

I am not sure about the last part, please check it with someone else!

d) if the force is removed, the bar will come to rest (ma=-8N)

a=-2m/s2 (therefore time taken to reach 0m/s using v=u+at =0.75s)

velocity at any time v=1.5-2t

induced Emf at any time t= BL(1.5-2t)

current at any time t=BL(1.5-2t)/R

P=i2R =(B2L2/R)(1.5-2t)2

=(16/3)(1.5-2t)2

Integrate this from 0 to 0.75, and that should be your answer!

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