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Current carrying bar in a magnetic field

  • Thread starter BOAS
  • Start date
  • #1
555
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Hello,

I am having trouble seeing what is going on here and would really appreciate a discussion about the situation.

1. Homework Statement

A bar of length ##L = 20 cm## and negliglible mass can slide over two conducting rails connected to a dc generator producing an emf ##V_0 = 6 V##, connected so as to produce a current as in the figure. The resistance of the bar is ##R = 0.08 Ω##, all other parts having negligible resistance. The bar is connected through a pulley to a body of mass ##m = 1.2 kg## (see figure attached). The system is immersed in a uniform magnetic field orthogonal to the rails, as in the figure, whose magnitude is ##B = 1 T##. The system is designed in such a way that, after a while, the body is pulled upward with a constant limiting speed of magnitude ##v_∞##. Compute, when the limiting speed is reached:

(a) the current flowing in the circuit;

(b) the magnitude ##\mathcal{E}## of the induced electromotive force;

(c) the magnitude v∞ of the limiting speed.

(d) Compute the value ##R_0## of the resistance of the bar corresponding to which the body does not move at all.

(e) The sliding bar is replaced with another bar having resistance ##R = 2R_0##, with ##R_0## the resistance computed in part (d). What is the numerical value of the limiting speed ##v_∞## in this case?

(f) Let us consider again the case ##R = 2R0##. Using the equation of motion for the mass ##m## and the equation for the current ##I##, determine the velocity ##v(t)## of the mass ##m##, assumed to be initially at rest. Express your answer in terms of ##v_∞## and ##g##.

Homework Equations




The Attempt at a Solution



The fact that the mass is moving upwards is really confusing me.

The weight attached to the pulley is acting to pull the bar towards the right.
The magnetic force on the current carrying bar acts to pull the bar to the left.
The fact that the bar is moving tells us that an EMF is induced that opposes this motion.

I think what I have is this ##(I_1 - I_2)LB = mg##, where ##I_1## is the current due to the battery and ##I_2## is due to the induced emf. The forces are equal due to no acceleration.

This suggests that a current flows with magnitude ##I = \frac{mg}{LB}## anticlockwise.

I haven't really managed to convince myself of this, so i'd really appreciate some help.

Thank you.
 

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Answers and Replies

  • #2
Hesch
Gold Member
922
153
This suggests that a current flows with magnitude I=mgLBI = \frac{mg}{LB} anticlockwise.
Yes, calculate I.

Now, I = ( V0 - ε ) / R

and

ε = dψ/dt , where ψ is the flux through the closed loop ( number of turns = 1 ).
 

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