Question

Find the terminal speed of the rod if it has mass m = 2.8 grams , length l = 21 cm , and resistance R = 0.0011 Ω . It is falling in a uniform horizontal field B = 0.052 T . Neglect the resistance of the U-shaped conductor.

Answer 1

Answer:

0.253 m/s

Explanation:

As the conductor falls down, the magnetic flux throug the coil formed by the conductor and the the rest of the circuit changes, therefore an electromotive force is induced in the rod; its magnitude is given by

$E=BvL$

where

B = 0.052 T is the strength of the magnetic field

v is the speed at which the rod is falling

L = 21 cm = 0.21 m is the length of the rod

Due to this electromotive force, a current is also induced in the rod and the circuit, and this current is given by

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

where

$R=0.0011 \Omega$ is the resistance of the rod

So the current is

$I=\frac{BvL}{R}$ (1)

At the same time, we know that a current-carrying wire in a magnetic field experiences a force, which is given by

$F_B = IBL$ (2)

where in this case:

I is the induced current given by eq(1)

B is the strength of the magnetic field

L is the lenght of the rod

Inserting eq(1) into (2), we find that the magnetic force on the rod is:

$F_B=\frac{BvL}{R}\cdot BL = \frac{B^2 L^2 v}{R}$

However, there is another force acting on the rod: the force of gravity, given by

$F_g=mg$

where

$m=2.8 g = 0.0028 kg$ is the mass of the rod

$g=9.8 m/s^2$ is the acceleration due to gravity

The falling rod will reach its terminal speed when its net acceleration becomes zero; this occurs when the net force on it is zero, so when the magnetic force is balanced by the force of gravity, so when

$F_B = F_g$

So

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

And solving for v, we find the terminal speed:

$v=\frac{mgR}{B^2L^2}=\frac{(0.0028)(9.8)(0.0011)}{(0.052)^2(0.21)^2}=0.253 m/s$