A bar of length L = 0.36 m is free to slide without friction on horizontal rails

Question

A bar of length L = 0.36 m is free to slide without friction on horizontal rails, as shown in the following figure. There is a uniform magnetic field B = 1.5 T directed into the plane of the figure. At one end of the rails there is a battery with emf epsilon = 12 V and a switch. The bar has mass 0.90 kg and resistance 5.0 Ohm, and all other resistance in the circuit can be ignored. The switch is closed at time t = 0. (a) Sketch the speed of the bar as a function of time. (b) Just after the switch is closed, what is the acceleration of the bar? (c) What is the acceleration of the bar when its speed is 2.0 m/s? (d) What is the terminal speed of the bar?

Explanation / Answer

a.

b

b. just after the switch is closed, current in the circuit = emf of the battery/resistance= 12/5= 2.4 A

force on the bar = F= current x length of conductor x B

= 2.4 x 0.36 x 1.5

= 1.296 N

acceleration of the bar= force on the bar/mass = 1.296 /0.9 = 1.44 m/s2

c. when speed is 2 m/s, back emf produced= magnetic field x length of bar x speed of bar

=1.5 x 0.36 x 2 = 1.08 V

so net emf in the circuit = 12 - 1.08 = 10.92 V

current in the circuit =10.92 / resistance =10.92/5 = 2.184 A

so force on the bar = current x length of bar x magnetic field

=2.184 x 0.36 x 1.5

= 1.17936 N

acceleration of the bar = force/mass

=1.17936 /0.9 = 1.31 m/s2

d.terminal speed is the speed where back e.m.f.= cell e.m.f.

so, 12 = magnetic field x length of bar x speed

so, 12 = 1.5 x 0.36 x speed

so, terminal speed of bar = 22.222 m/s2

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