A metal rod of massm= 100 g and lengthL= 1.0 m rests on two fixed,frictionless,

Question

A metal rod of massm= 100 g and lengthL= 1.0 m rests on two fixed,frictionless, conducting rails, as seen from above in the diagram.SwitchSis closed at timet= 0, allowing the batteryVb= 50 V to generate currentthrough the top rail, down through the rod, and then through the other railback to the battery.The wires have a combined resistanceR= 2andthey are immersed in a uniform magnetic field of strengthB= 0.02 Tpointing downward (as shown)

What is the current in the circuit at timet= 2.0 s?

What is the force on the rod at timet= 2.0 s?

What is the velocity of the rod at timet= 2.0 s?

Explanation / Answer

If the rod is moving with a velocity V

Then changeof flux rate = BLv

Soinduced EMF = Blv

Now from loop law 50 - BLv = I * R

I = (50 - BLv) /R

F = B* I * L = B*L( 50 - B*L*v)/R

Now Force = change in momentum

So m dv/dt = B*L( 50 - B*L*v)/R

dv/( 50 - BLv) = BL/Rm * dt

(-1/BL) ln (50 - BLv) = BL t / Rm

Apply limits from 0 to V and O to t

ln (50 - BLV) /50 = - B^2 L^2 t /Rm

50 - BLV = 50 * e ^- B^2L^2 t/Rm

v = - 50 ( e ^ - B^2 L^2 t - 1)/ BL

Calculate v at 2 sandsubstitute to get force and current

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