THE OSCILLATING MAGNET

INTRODUCTION

As the name of the project suggests " THE OSCILLATING MAGNET", in this project we shall obtain the solution of the Differential Equation which is similar to the Differential Equation of Simple Harmonic Motion i.e., d^2x=dt^2 = -kx , where k is a constant.

Here, by substituting the value of time in the solution obtained by solving the differential equation, we can obtain the position of the oscillating magnet

ARRANGEMENT OF THE APPARATUS

When a current carrying coil and a bar magnet of pole strength(p) are arranged such that the current carrying coil is fixed and the bar magnet can move vertically on the top of the axis of the current carrying coil, oscillatory motion of the bar magnet is observed.

Here, we are assuming the negligible air resistance near the oscillating magnet.By assuming negligible air resistance, the damping effect can be reduced.This assumption leads to the simplicity. Despite the simplicity of the setup, it reveales few interesting relations.

AN ANALYTICAL SOLUTION

In this setup the bar magnet floats in the air because the Force of Gravity and the Magnetic Force acting on the bar magnet

due to the current carrying coil is balanced at the equilibrium position.

The magnitude of the magnetic eld on the axis of the current

carrying coil having N turns is given by

B_z =( u_0NIR^2)/2(R^2+x^2)^(3/2)

[2] Assuming R<<x,

Our equation becomes

B_z = (u_0NIR^2)/2x^3

Here, u_0 = 4*10^(-7) N/A^2

I=Current

owing through the loop

N=Number of turns of the loop

R=Radius of the coil

x=Distance from the plane of the loop to the midpoint of the bar magnet

The Magnetic Force acting on the bar magnet is given by

F_m = p*B_z

where p=Pole Strength of the bar magnet

Thus, our equation becomes

F_m =( u_0NIR^2p)/2x^3

The Force of Gravity acting downwards on the bar magnet is given by

F_g = mg

At the equilibrium position,

mg=(u_0NpIR^2)/2x^3

Now, if we give a small displacement(x_0) downwards, then the equation of the resultant Force is given by

F={pu_0IR^2/2}[{ 1/(x-x0)^3}-{1/x^3}]

F=3u_0pIR^2x_0/2x^4

But F = ma

a=3u_0pIR^2x_0/2mx^4

d^2x_0/dt^2 = kx_0

where k=3u0pIR^2/2mx^4

Now, the above equation is a Second Ordered Linear Ordinary Differential Equation with Constant Coefficient

we get our final solution of the differential equation as

x(t)=c1e^(k^(1/2)t) + c2e^(-(k)^1/2t)

CONCLUSION

The vertical oscillations of bar magnet yield an interesting relation despite its simplicity.We studied small amplitude

oscillations, and found that the position of the bar magnet at a particular time can be predicted.

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