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A mini rail gun made of a copper rod of mass m and radius r rests on two parallel copper rails that are a distance L apart and of length d. A magnetic field of magnitude B is directed perpendicular to the rod and the rails. When current is applied to the parallel copper rails, the rod rolls along the rails without slipping. If the rod starts from rest, what is the speed of the rod as it leaves the rails?

Respuesta :

Answer:

Explanation:

Magnetic Force on the rod  F = Bi L

Work done by this force = F X d

=  Bi Ld

This energy is converted into both rotational and linear kinetic energy

= 1/2 I ω² + 1/2 mv²

= 1/2 x 1/2 m r²ω²+ 1/2 mv²

= 1/4 m v² + 1/2 mv²

= 3/4 mv²

So according to conservation of energy

Bi Ld =  3/4 mv²

v = [tex]\sqrt{\frac{4BiLd}{3m} }[/tex]

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