Question

A rod of m= 1.3 kg rests on two parallel rails that are L = 0.42 m apart. The rod carries a current going between the rails (bottom to top in the figure) with a magnitude 1 = 2.6 A. A uniform magnetic field of magnitude B = 0.35 T pointing upward is applied to the region, as shown in the graph. The rod moves a distance d=1.25 m. Ignore the friction on the rails. † † † Ē I Otheexpertta.com A Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails.Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails. Assume the current through the rod is constant at all times.

Answer 1

Answer:

The final speed of the rod is 0.86 m/s.

Explanation:

Given that,

Mass of rod = 1.3 kg

Distance between rail= 0.42 m

Current = 2.6 A

Magnetic field = 0.35 T

Distance = 1.25 m

We need to calculate the acceleration

Using formula of magnetic force

$F= Bil$

$ma=Bil$

$a=\dfrac{Bil}{m}$

Put the value into the formula

$a=\dfrac{0.35\times2.6\times0.42}{1.3}$

$a=0.294\ m/s^2$

We need to calculate the final speed of the rod

Using equation of motion

$v^2-u^2=2as$

Put the value in the equation

$v^2=2\times0.294\times1.25$

$v=0.86\ m/s$

Hence, The final speed of the rod is 0.86 m/s.

Answer 2

The final speed of the rod is 0.86 m/s.

What is Speed?

This refers to the rate of change of the position of an object in a specified direction.

The Parameter

• Mass of rod = 1.3 kg
• Distance between rail= 0.42 m
• Current = 2.6 A
• Magnetic field = 0.35 T
• Distance = 1.25 m

To calculate the acceleration

We use the formula of magnetic force

a=Bil/m

a= (0.35 x 2.6 x 0.42)/1.3

a= 0.294m/s^2

Then the final speed of the rod

We use the equation of motion

v^2 - u^2= 2as

=> v^2= 0.86m/s

Read more about speed here:

https://brainly.com/question/6504879