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nawand

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## Homework Statement

A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails fig.1 that are the d=12.0 cm apart and L=45.0 cm long. The rod carries a current of I=48.0 A in the direction shown and rolls along the rails without sleeping. A uniform magnetic field of magnitude 0240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?

## Homework Equations

0+0+F

_{b}*L*cos θ = (mv

^{2})/2 +(Iω

^{2})/2

## The Attempt at a Solution

The rod gains energy from the rest, so begin kinetic energy and rotational energy are zero. The only source of energy rod gains is magnetic filed.

I = (mr

^{2})/2

K=(mv

^{2})/2

w=v/r

That is right side of equations becomes:

(mv

^{2})/2 + 1/2 (mv

^{2})+1/2(1/2 mr

^{2}) * (v

^{2}/r

^{2})

F

_{b}* L = 3/4 mv

^{2}

Where

F

_{b}= I * d * B

etc

v=1.07 m/s

Because the force move the rod parallel to the rails along x axis, the dot product will be:

E=F

_{b}* L * cos θ = F

_{b}*L*1=F

_{b}L

The question is how do we get the value of energy exerts magnetic field on the rod?

That is force time Length of rails (F

_{b}times L)?

For the electric field is E = qF

_{e}