Question

A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart. A 5.5×10^−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.
Express your answer using two significant figures.

Answer 1

Answer:

Terminal speed, v = 6901.07 m/s

Explanation:

It is given that,

Mass of the horizontal bar, m = 30 g = 0.03 kg

Length of the bar, l = 13 cm = 0.13 m

Magnetic field, $B=5.5\times 10^{-2}\ T$

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,  

Force of gravity = magnetic force

$mg=ilB$..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

$E=Blv$, v is the speed of the bar

$iR=Blv$

$i=\dfrac{Blv}{R}$

Equation (1) becomes,

$mg=\dfrac{B^2l^2v}{R}$

$v=\dfrac{mgR}{B^2l^2}$

$v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}$

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.