Multiple choice Q: Mangetic fields and Spring Constant

[User1265]

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.

My attempt at a solution: Is my logic accurate/correct, and is my answer correct?

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?

 

[Doc Al]

Looks good to me.

 

[kuruman]

I think answer (B) is incorrect. The rod may have zero resistance but it doesn't have zero mass. Also, I don't know what k_eff means. When the current is switched on, the magnetic force $ILB$ is constant and its effect is to shift the equilibrium position of the hanging rod. If anything is effective, it is the acceleration of gravity.

I would write the force equilibrium equations for (a) zero current, (b) current = $I$ and (c) current = $\frac{I}{2}$ and express the difference in equilibrium positions c-a in terms of the difference b-a which is known to be $y$.

 

[Charles Link]

I think answer (B) is incorrect. The rod may have zero resistance but it doesn't have zero mass. Also, I don't know what k_eff means. When the current is switched on, the magnetic force $ILB$ is constant and its effect is to shift the equilibrium position of the hanging rod. If anything is effective, it is the acceleration of gravity.

I would write the force equilibrium equations for (a) zero current, (b) current = $I$ and (c) current = $\frac{I}{2}$ and express the difference in equilibrium positions c-a in terms of the difference b-a which is known to be $y$.

I think they are already considering the weight of the rod to be balanced by the springs, in both cases. They simply ask for the (additional) displacement as the switch is closed.

 

[Doc Al]

I think answer (B) is incorrect. The rod may have zero resistance but it doesn't have zero mass.

Sure, the rod has mass but that doesn't change the answer. It starts out with no current at some equilibrium point. All displacements are with respect to that point.

 

[kuruman]

I think they are already considering the weight of the rod to be balanced by the springs, in both cases. They simply ask for the (additional) displacement as the switch is closed.

Sure, the rod has mass but that doesn't change the answer. It starts out with no current at some equilibrium point. All displacements are with respect to that point.

Right. With the current off, the equilibrium position $x_{\text{0A}}$ is such that $kx_{\text{0A}}=mg$. We turn the current on and the new equilibrium position is $x_{\text{0B}}$ such that $y=x_{\text{0A}}-x_{\text{0B}}.$ Finally we turn the current off, double the resistance and turn the current back on. The third equilibrium position is $x_{\text{0C}}$. We are asked to find $x_{\text{0A}}-x_{\text{0C}}$ in terms of $y.$

Write the force equilibrium equations, find the three equilibrium positions and see what you get for the differences.

 

[Doc Al]

Write the force equilibrium equations, find the three equilibrium positions and see what you get for the differences.

Measure displacements from the initial position, where the rod hangs with no current through it. At that point, the net force on the rod is zero (spring forces balance gravity). Now flick the switch to add current and the rod moves to its new equilibrium point, a distance y from the start. The spring force changes and there's a magnetic force: These forces must balance, giving: $$y = \frac{ILB}{2k}$$.

Now redo it with spring constant doubled and half the current. Again, measuring from the zero-current starting point (a different point than before, but so what), you get the new displacement: $$y_2 = \frac{y}{4}$$.

Nothing wrong with doing it as you suggest, but you'll get the same answer. Or did I miss something?

 

[Charles Link]

Probably a good exercise for a student to work the problem of a spring with the additional gravitational force that simply introduces an offset, so that the system considered with the other forces still has the same spring constant.

 

[kuruman]

Ooops! I missed the part that says that the spring constant is doubled. Sorry guys.