Suppose that a conducting rod of mass 0.059 kg and resistance 4.990 ohms rests o

Question

Suppose that a conducting rod of mass 0.059 kg and resistance 4.990 ohms rests on two frictionless and resistanceless rods a distance 0.20m apart in a uniform magnetic field B = 0.069T perpendicular to the rails and the rod, as shown in the figure above. A source of emf is connected to points a and b. At t = 0, the rod is at rest and the current is 4.5 A and maintained at that value. How fast is the rod moving at t = 6.47 s? (Ignore the back emf)

Please show calculation, thanks.

1. 3.239 m/s

2. 6.810

3. 14.32

4. 30.10

5. 63.29

6. 4.697

7. 9.874

8. 20.76

9. 43.65

10. 91.77

Explanation / Answer

The force on the rod from the magnetic field is found by the formula F = BIL

The force will move the rod according to Newton's Second Law such that F = ma, so we can set those equal to each other

BIL = ma

Solve for the acceleration

(.069)(4.5)(.2) = (.059)(a)

a = 1.052 m/s2

Next apply the formula vf = vo + at to find the final velocity (initial velocity is zero)

vf = (0) + (1.052)(6.47)

vf = 6.810 m/s which is choice number 2

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