Question

Give a rail gun shown below with B-3.0T and rail spacing of 25 cm. A discharging capacitor supplies a current of 100 kA. 5) a. What is the acceleration if the bar has a mass of 0.25 kg. [3 x 10s m/s?] h. What is the bars speed at the end of a 2.75 m long track [1.2 km/s ]

Answer 1

Step-by-step explanation:

It is given that,

Magnetic field, B = 3 T

Length of the spacing, L = 25 cm = 0.25 m

Current,
`I=100\ kA=100* 10^3\ A=10^5\ A`

Mass of the bar, m = 0.25 kg

(a) Magnetic force is given by :

`F=ILB`

`F=10^5\ A* 0.25\ m* 3\ T`

F = 75000 N

Using second law of motion,

F = m × a

`a=(F)/(m)`

`a=(75000\ N)/(0.25\ kg)`

`a=300000\ m/s^2`

`a=3* 10^5\ m/s^2`

(B) Let v is the speed of the bars at the end of long track. Initial speed of the bar is 0. Using third equation of motion as :

`v^2-u^2=2as`

`v^2=2as`

`v=√(2as)`

`v=√(2* 300000\ m/s^2* 2.75\ m)`

v = 1284.52 m/s

or

v = 1.2 km/s

So, the speed of the bar at the end is 1.2 km/s. Hence, this is the required solution.