A metal rod is forced to move with constant velocity = 75.0 cm/s along two parallel metal rails, connected with a strip of metal at one end, as shown in the figure. A magnetic field B = 0.35 T points out of the page. The rails are seperated by 35.0 cm. a)What emf is generated?  b)If the rod has a resistance of 17.0 Ω and the rails and connectors have negligible resistance, what is the current in the rod? c)At what rate is energy being tranferred to thermal energy?

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A metal rod is forced to move with constant velocity = 75.0 cm/s along two parallel metal rails, connected with a strip of metal at one end, as shown in the figure. A magnetic field B = 0.35 T points out of the page. The rails are seperated by 35.0 cm.

a)What emf is generated? 

b)If the rod has a resistance of 17.0 Ω and the rails and connectors have negligible resistance, what is the current in the rod?

c)At what rate is energy being tranferred to thermal energy?

L
V
B
Transcribed Image Text:L V B
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Hello. Every single one of the answers given here is wrong. 

(8c30p31) A metal rod is forced to move with constant velocity = 75.0 cm/s along two parallel metal rails
of metal at one end, as shown in the figure. A magnetic field B = 0.35 T points out of the page. The rails
cm.
What emf is generated?
0.0672 V
Check the units for MKS and emf=v | B
Submit Answer Incorrect. Tries 1/4 Previous Tries
If the rod has a resistance of 17.0 2 and the rails and connectors have negligible resistance, what is the c
3.963e-3 A
Check the units for MKS and i=emf/R
Submit Answer Incorrect. Tries 1/4 Previous Tries
At what rate is energy being tranferred to thermal energy?
2.67e-4 W
Check the units for MKS and power i^2 R
Submit Answer Incorrect. Tries 1/4 Previous Tries
B
This discussion is closed.
Expert Solution
Step 1
Given,
v = 75 cm/s
B = 0.35 T
L = 35 cm
Step 2
a) Emf generated is given by
E = BLV = (0.35 T)*(0.35 m)*(0.55 m/s)
E = 0.0673 V
b) Given, R = 17.00
By Ohm's Law, induced current is
1 = E/R = 0.0673 V/ 17.00
1 = 3.963*10-³ A
By Lenz's Law, current is clockwise.
c) Rate of energy to thermal energy
P = 1²R = (3.963*10-³ A)² * (17)
P = 2.67*10-4 W
A metal rod is forced to move with constant velocity
⇓ C
Transcribed Image Text:(8c30p31) A metal rod is forced to move with constant velocity = 75.0 cm/s along two parallel metal rails of metal at one end, as shown in the figure. A magnetic field B = 0.35 T points out of the page. The rails cm. What emf is generated? 0.0672 V Check the units for MKS and emf=v | B Submit Answer Incorrect. Tries 1/4 Previous Tries If the rod has a resistance of 17.0 2 and the rails and connectors have negligible resistance, what is the c 3.963e-3 A Check the units for MKS and i=emf/R Submit Answer Incorrect. Tries 1/4 Previous Tries At what rate is energy being tranferred to thermal energy? 2.67e-4 W Check the units for MKS and power i^2 R Submit Answer Incorrect. Tries 1/4 Previous Tries B This discussion is closed. Expert Solution Step 1 Given, v = 75 cm/s B = 0.35 T L = 35 cm Step 2 a) Emf generated is given by E = BLV = (0.35 T)*(0.35 m)*(0.55 m/s) E = 0.0673 V b) Given, R = 17.00 By Ohm's Law, induced current is 1 = E/R = 0.0673 V/ 17.00 1 = 3.963*10-³ A By Lenz's Law, current is clockwise. c) Rate of energy to thermal energy P = 1²R = (3.963*10-³ A)² * (17) P = 2.67*10-4 W A metal rod is forced to move with constant velocity ⇓ C
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