Answer :
Explanation:
It is given that,
Magnetic field, B = 3 T
Length of the spacing, L = 25 cm = 0.25 m
Current, [tex]I=100\ kA=100\times 10^3\ A=10^5\ A[/tex]
Mass of the bar, m = 0.25 kg
(a) Magnetic force is given by :
[tex]F=ILB[/tex]
[tex]F=10^5\ A\times 0.25\ m\times 3\ T[/tex]
F = 75000 N
Using second law of motion,
F = m × a
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{75000\ N}{0.25\ kg}[/tex]
[tex]a=300000\ m/s^2[/tex]
[tex]a=3\times 10^5\ m/s^2[/tex]
(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 :
[tex]v^2-u^2=2as[/tex]
[tex]v^2=2as[/tex]
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2\times 300000\ m/s^2\times 2.75\ m}[/tex]
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.