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A conductor lies along the Z-axis at    -1.5\leq z< 1.5 \: m and carries a fixed current of 10.0 A in  - \hat{a_{z}} direction (see figure).

for a field \vec{B}=3.0\times 10^{-4}\: e^{-0.2x}\: \: \hat{a_{y}}   T ,Find the power required to move the conductor at constant speed to x=2.0 m,

y = 0 m in 5\times 10^{-3}s.Assume parallel motion along the x-axis.

Option: 1

1.57 W


Option: 2

2.97 W


Option: 3

14.85 W


Option: 4

29.7 W


Answers (1)

best_answer

As we discussed in @

w=\int_{0}^{2}F.dx=\int_{0}^{2}3\times 10^{-4}e^{-0.2x}\times 10\times3dx

w=9\times 10^{3}\int_{0}^{2}e^{-0.2x}dx=\frac{9\times 10^{-3}}{0.2}\left [ e^{-0.2\times 2}+1 \right ]

\beta =3.0\times 10^{-4}e^{-0.2x}= \frac{9\times 10^{-3}}{0.2}\times \left [ 1-e^{-0.4} \right ]=\frac{9\times 10^{-3}\times \left ( 0.33 \right )}{0.2}= \frac{}{2.97\times 10^{-3}J}{0.2}\\\therefore P= \frac{w}{f}= \frac{2.97\times 10^{-3}}{\left ( 0.2 \right )\times 5\times 10^{-3}}= 2.97w

Posted by

Riya

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