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Lubov Fominskaja [6]
3 years ago
12

An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 1.95 mT. If the speed of the

electron is 1.44 107 m/s, determine the following. (a) the radius of the circular path _____ cm (b) the time interval required to complete one revolution ____ s
Physics
1 answer:
zvonat [6]3 years ago
4 0

Answer: a) 42 *10^-3 m= 42 mm; b) 10.40 *10^-6 s=10.40 μs

Explanation: In order to response this problem we have to consider the Newton law for the circular movement,

Fm=m*ac where Fm is the magnetci force and ac the centripetal acceleration which is equal to v^2*R ( the raduis of teh curcular trajectory)

Fm=e*v*B considering that v and B are perpendicular

then we have:

e*v*B=m*v^2/R so

R=m*v/(e*B)= 9.1*10^-31*1.44*10^7/(1.6*10^-19*1.95*10^-3)= 42 mm

Then to calculate the time to complete one revolution ( period)

we know that ω=v*R and T= 2π/ω

then we have:

ω=1.44*10^7*42*10^-3=604.8 *10^3 rad/s

T=2*π/604.8 *10^3 = 10.4 μs

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There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

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Its cell potential would be equal to 0.339 - (-0.130) = \rm 0.469\; V.

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell}),

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  • n is the number moles of electrons transferred for each mole of the reaction. In this case the value of n is 2 as in the half-reactions.
  • F is Faraday's Constant (approximately 96485.33212\; \rm C \cdot mol^{-1}.)

\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}.

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