Question

A 60 g horizontal metal bar, 15-cm-long, is free to slide up and down between two tall, vertical metal rods that are 15 cm apart. A 5.0×10−2 T magnetic field is directed perpendicular to the plane of the rods. The bar is raised to near the top of the rods, and a 1.2 Ω resistor is connected across the two rods at the top. Then the bar is dropped.What is the terminal speed at which the bar falls? Assume the bar remains horizontal and in contact with the rods at all times.

Answer 1

Our values given are:

$m = 60 g=0.060kg\\L = 15 cm=15*10^{-2}m\\B = 0.05 T\\R = 1.2 \Omega$

For the second Newton Law we have,

$F = mg (1)$

And for the Faradays equation we have that

$\epsilon_{emf} = BLv (2)$

Where,

B= Magnetic field

L = Lenght

v = velocity

However for the Ohm's Law we have that

$\epsilon_{emf}=V=I*R (3)$

V = Voltage

I = Current

R = Resistance

Equating equation (2) and (3)

$IR = BLv$

$I = \frac{BLv}{R}$

The magnetic force on a charged particle depends on the relative orientation of the particle's velocity and the magnetic field. And it is defined as,

$F = IBL$

Replacing the previous value of the current,

$F = \frac{B^2L^2v}{R}$

And replacing the value of the force given by Newton we have,

$mg = \frac{B^2L^2v}{R}$

Re-arrange to find the velocity,

$v=\frac{mgR}{B^2L^2}$

We can now replace the values of this problem,

$v = \frac{(0.06)(9.8)(1.2)}{(5*10^{-2})^2(15*10^{-2})^2}$

$v = 12544m/s$