A conducting bar with length L = 0.36 m, mass m = 0.90kg, resistance R = 5.0 Ohm

Question

A conducting bar with length L = 0.36 m, mass m = 0.90kg, resistance R = 5.0 Ohm moves without friction on metal rails. Metal rails, the bar along with the battery (epsilon = 12.0 V) and switch S form a closed , circuit as shown in the Figure. A uniform magnetic field with magnitude B = 1.50T is directed into the plane. Initially, at t = 0 the bar is at rest, What is the acceleration of the bar just after the switch is S closed? What is its acceleration when its speed is 2.0 m/s? Derive an expression for the terminal speed of the bar. What is the terminal speed of the bar?

Explanation / Answer

A) force on wire is F = B*I*L*sin(90)..

since F = ma = B*I*L..

current I = e/R =12/5

accelaration a = B*I*L/m = 1.5*(12/5)*0.36/0.9=1.44 m/s^2...

B) accelaration is constnt a = 1.44 m/s^2...

C) induced emf e = B*L*v..

v is terminal speed...

v = e/(B*L)....

D) v = e/(B*L) = 12/(1.5*0.36)=22.23 m/sec

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