# Spring in an electrical field

1. Homework Statement
A rigid rod of mass m and length l is suspended from two identical springs of negligible mass such that one spring is attached at either end of the rod. The upper ends of the springs are fixed in place and the springs stretch a distance d under the weight of the suspended rod. I did manage to solve this and find the k=(ma)/(2d) however the question will continue to say If the springs are a part of a circuit connected to a battery with negligible internal resistance and E with a Resister with resistance R and a switch(there is a picture however it is not of much use) Then the rod will move by delta d down from the springs. The total resistance of the circuit is R (the same as the one resister, assuming the wires have negligible resistance.) The rod is in uniform magnetic field directed perpendicular to the field the upper ends of the springs remain fixed in place and the switch is closed. When the system comes to equilibrium the rod is lowered by the prior mentioned delta d.
a)What is the direction of the magnetic field according to the coordinate axis with x to the right y up and z out of the page
b)Determine the magnitude of the magnetic field in terms of: m, d,delta d, E (epsilon) R and fundamental constants.
c) when the switch is suddenly opened the rod oscillates. For these oscillations determine the following in terms of d delta d and fundamental constants, the period (T) and the max speed of the rod.

Relevant equations
F=qvb
F=IBl

3. The Attempt at a Solution
I got a using the right hand rule however beyond there I am clueless please help. This is a high school AP physics class with no calculus please, thank you much.

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b)Determine the magnitude of the magnetic field in terms of: m, d,delta d, E (epsilon) R and fundamental constants.
c) when the switch is suddenly opened the rod oscillates. For these oscillations determine the following in terms of d delta d and fundamental constants, the period (T) and the max speed of the rod.
I am presuming that the problem statement means that the rod descends by an additional amount delta_d, since the weight of the rod will still stretch the springs by d from their "resting lengths" with the magnetic force acting as well as gravity.

So you know that the spring constant is k = mg/2d and that the magnetic force on the rod must equal the weight of the rod, since that stretches the springs just as much as the rod's weight did. That magnetic force acting on the current in the rod is

F_mag = I·B·l .

This force will stretch the springs by the additional delta_d . Hooke's Law will give you the necessary relationship and you have an expression for the spring constant. Ohm's Law will relate I, E and R. I believe this should let you find the requested expression for the magnetic field strength B (part b).

For part (c), the removal of the current eliminates the magnetic force. The rod is now just a mass m in a harmonic oscillator with spring constant k. What is the period of such an oscillator? You would then replace k with your expression to get the expression requested for T. (Why does delta_d not appear there?)

The kinetic and potential energies in the oscillator are what you want to look at for the last question. The springs are stretched by delta_d due to the magnetic force and are then released. What is the potential energy in a stretched spring? How does that compare with the maximum kinetic energy the oscillating rod will have? You can get an expression for the maximum speed of the rod in terms of the added amount the springs were stretched before release. Replacing k again will give you the requested expression for the rod's maximum speed.

In part c, to calculate the period of oscillation and the maximum speed of the rod, would the effective spring constant be 2k (mg/d) since the two springs are in parallel?