Work done on an e-

[arc5005]

TBR Physics II Book Diagnostic Eval #2.1 Passage II:

Q9: How much work is done on an electron in moving it across the bar in Figure 1?

a. 0
b. BLv/e
c. evLB
d. evB/L

Answer/Explanation:

C) evLB
The work required to push an e- from 1 end of the moving rod to the other is equal to the potential energy change:
W = ΔU = qΔV
From the passage, we are told that the induced emf accross the ends of the bar is given by:
ε = BLv.
The work required to move an electron is given by: W = BLve

My question:

where did the e come from? Can anyone explain this a little better please? is epsilon supposed to be the V (voltage)?

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[PhilzCoffeeAddict]

Okay so lets work backwards a bit. You're trying to find the work an electron does. The equation for work (also can be energy) for an electron is W =qV, where W = work, q = charge of electron, V = voltage.

ε = emf = voltage. Electron motive force is the potential that drives the electron moving. So through plugging in ε for voltage. We get.

W=qV = q(BLv) = eBLv = evLB

The e is just -q but in the question, the negative sign is missing. So you are correct ε is voltage/electron motive force. e = q (really -q, but for this question its just q).

Does this make sense?

 

[arc5005]

Okay so lets work backwards a bit. You're trying to find the work an electron does. The equation for work (also can be energy) for an electron is W =qV, where W = work, q = charge of electron, V = voltage.

ε = emf = voltage. Electron motive force is the potential that drives the electron moving. So through plugging in ε for voltage. We get.

W=qV = q(BLv) = eBLv = evLB

The e is just -q but in the question, the negative sign is missing. So you are correct ε is voltage/electron motive force. e = q (really -q, but for this question its just q).

Does this make sense?

yes thank you!