A 40 g horizontal metal bar, 14 cm long, is free to slide up and down between tw

Question

A 40 g horizontal metal bar, 14 cm long, is free to slide up and down between two tall, vertical metal rods that are 14 cm apart. A 5.5×102 T magnetic field is directed perpendicular to the plane of the rods. The bar is raised to near the top of the rods, and a 1.0 resistor is connected across the two rods at the top. Then the bar is dropped.

What is the terminal speed at which the bar falls? Assume the bar remains horizontal and in contact with the rods at all times.

Express your answer using two

Explanation / Answer

Use the expression -

F = mg

and

F = ILB

equalize the both we have -

I = mg/(LB) -----------------------------------------------------(a)

again -

Energy = Fv = mgv = I^2R

=> v = I^2R/(mg)

put the value of 'I' from (a)-

v = (mg/(LB))^2R/mg = mgR/(LB)^2
given that -
B = 0.055 T, g = 9.81 m/s^2, m = 40 g (0.04 kg), L = 14 cm (0.14 m), R = 1 ohm

So -  
v = 0.04*9.81*1/(0.14*0.055)^2 = 6611 m/s

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