- #1

User1265

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- 1

- Homework Statement
- The diagram shows a metal rod suspended in a magnetic field by two vertical conducting springs. The cell and rod have negligible resistance. When the switch S is closed the effect of the magnetic field is to displace the rod vertically a distance y. When both the spring constant and electrical resistance of each spring is doubled, closing the switch would now cause the rod to be displaced a distance.......

- Relevant Equations
- F= ILB

F=kx

y is the notation I use for displacement in the given context.

I consider the forces acting to be: Restoring forces in springs parallel, and Force of the current-carrying conductor in the Magnetic field. I imagine a vertical displacement of y upwards ( direction determined by Fleming's Left-Hand Rule, Field going into the plane away from the observer, Conventional current left to right, the force exerted on the rod, initially at rest, is upwards ).

(1) I then consider the electrical resistance doubled. Since Voltage is unchanged, a doubling of the total resistance (in series) in the circuit of parallel springs, will half the current in the circuit and thus half the current through the rod.

So the Force, F, acting upwards on the current-carrying wire is *1/2 already, as Force on current-carrying wire=ILB, and I*1/2.

Fcurrent carrying wire ∝ I, I *1/2 ⇒ Fcurrent carrying wire*1/2Then I imagine, that Force exerted due to the current-carrying wire, will compress the springs upwards until at y displacement where equilibrium is reached between the two equal internal elastic forces of the spring and the Force of the current-carrying conductor.

The compressing force, in this case, is the force of the current-carrying conductor, which is proportional to the displacement the spring compressed by, is halved so the displacement is halved. F*1/2 ---> so y/2 (for an unchanged stiffness constant of k)

Fcurrent carrying wire is Fcompressing

Fcompressing ∝ y , Fcurrent carrying wire*1/2 ⇒ y *1/2

(2)Considering now the isolated effect of doubling of the stiff/spring constant, I imagine that the metal rod would a Force F reach equilibrium with the elastic forces at half the distances, as the parallel springs have doubled in stiffness. Since I equated F current-carrying wire to the Fcompressing on the spring, then for this unchanged Force F - when k of each spring is doubled, effective spring constant doubled, the displacement y is halved.

1/k eff ∝ y (unchanged Fcurrent-carrying wire)

k eff new = 2 * k eff inital

so k *2 ⇒ y*1/2Combining effects (1) and (2) on displacement is a quartering effect on displacement y , so Answer is B) y/4Is my logic correct?