jlo- 4x x 10 T-m/A 8.85 x 10- CNem Power = Force. velocity 1-emfR F-IIXB t-EI .

Question

jlo- 4x x 10 T-m/A 8.85 x 10- CNem Power = Force. velocity 1-emfR F-IIXB t-EI . Biv 1. (15 pts) A conducting bar moves along fri tonless conducting rails connected to a 4.00- resistor as shown in the fig- ure. The length of the bar is 1.60 m and a uniform magnetic field of 2.20 T is applied perpendicular to the paper pointing outward, as shown. a. What is the applied force required to move the bar to the right with a constant speed of 6.00 m/s? b. At what rate is energy dissipated in the 4.00 resistor?

Explanation / Answer

a)

flux = B*A

A = area = L*dx

flux = B*L*dx

induced emf = rate of change in flux

e = B*L*dx/dt

induced current = i = e/R

R = resistance

magnetic force on the rod = Fb = i*L*B = e*L*B/R = B*L*v*L*B/R = B^2*L^2*v/R

Fapplied = Fb

Fapplied = (2.2^2*1.6^2*6)/4 = 18.6 N

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b)

i = e/R = B*L*v/R = (2.2*1.6*6)/4 = 5.28 A

power = i^2*R = 5.28^2*4 = 111.5114 W

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